# 1.1 ANALYTICAL GEOMETRY II

EQUATION OF CIRCLE

Definition:

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant.   The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.

Equation of the circle with centre (h, k)  and radius ‘r’ units.

CP = r

$\sqrt{(x\ -\ h)^2\ +\ (y\ -\ k)^2}\ =\ r\ (Using\ distance\ formula)$
$(x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2$

Note:

The equation of the circle with centre (0, 0 ) and radius ‘r’ units is  x+ y2  = r2

$\color {purple} {Example\ 1\ .}\ \color {red} {What\ is\ the\ equation\ of\ the\ circle}\ with\ centre\ at\ origin\ and\ radius\ 1\ unit?\ \hspace{15cm}$
$\color {blue} { Soln:}\ \hspace{20cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2$
$Given\ h\ =\ 0,\ k = 0\ and\ r\ =\ 1$
$(x\ -\ 0)^2\ +\ (y\ -\ 0)^2\ =\ 1^2$
$x^2\ +\ y^2\ =\ 1$
$The\ equation\ of\ the\ circle\ is\ \boxed{x^2\ +\ y^2\ =\ 1}$

$\color {purple} {Example\ 2\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ with\ centre\ (-5, 7)\ and\ radius\ 5\ units.\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ centre\ =\ (-5, 7)\ \hspace{4cm}\ and\ radius\ =\ 5\ \hspace{6cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2$
$Given\ h\ =\ -\ 5,\ k = 7\ and\ r\ =\ 5$
$(x\ +\ 5)^2\ +\ (y\ -\ 7)^2\ =\ 5^2$
$x^2\ +\ 10\ x\ +\ 25\ +\ y^2\ -\ 14y\ +\ 49\ =\ 25\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 10\ x\ -\ 14y\ +\ 25\ +\ 49\ -\ 25\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 10\ x\ -\ 14y\ +\ 49\ =\ 0\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 10\ x\ -\ 14\ y\ +\ 49\ =\ 0}\ \hspace{5cm}$

General equation of the circle:     x2  +   y2  + 2gx  + 2fy  + c = 0

Centre = (-g , -f )         and    radius     r =  √( g2  + f2 – c)

$\color {purple} {Example\ 3\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ x^2\ +\ y^2\ +\ 2\ x\ +\ 2\ y\ -\ 7\ =\ 0\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ x^2\ +\ y^2\ +\ 2\ x\ +\ 2\ y\ -\ 7\ =\ 0\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ 2\ \hspace{3cm}\ 2\ f\ =\ 2\ \hspace{3cm}\ c\ =\ -7$
$g\ =\ 1\ \hspace{3cm}\ f\ =\ 1\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{(g^2\ +\ f^2\ -\ c)}$
$centre\ =\ (-\ 1,\ -\ 1)\ \hspace{4cm}\ r\ =\ \sqrt{(1^2\ +\ 1^2\ +\ 7)}$
$\hspace{6cm}\ r\ =\ \sqrt{(1\ +\ 1\ +\ 7)}$
$\hspace{6cm}\ r\ =\ \sqrt{9}\ =\ 3$
$\fbox{centre = (- 1, – 1) r = 3}$

$\color {purple}{Example\ 4\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ x^2\ +\ y^2\ -\ 4\ x\ +\ 8\ y\ -\ 5\ =\ 0\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ x^2\ +\ y^2\ -\ 4\ x\ +\ 8\ y\ -\ 5\ =\ 0\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ -\ 4\ \hspace{3cm}\ 2\ f\ =\ 8\ \hspace{3cm}\ c\ =\ -\ 5$
$g\ =\ -\ 2\ \hspace{3cm}\ f\ =\ 4\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}$
$centre\ =\ (2,\ -\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{(-\ 2)^2\ +\ 4^2\ +\ 5}$
$\hspace{6cm}\ r\ =\ \sqrt{4\ +\ 16\ +\ 5}$
$\hspace{6cm}\ r\ =\ \sqrt{25}$
$\fbox{centre = (2, – 4) r = 5}$

$\color {purple} {Example\ 5\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ x^2\ +\ y^2\ -\ 8\ y\ +\ 3\ =\ 0\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ x^2\ +\ y^2\ -\ 8\ y\ +\ 3\ =\ 0\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ 0\ \hspace{3cm}\ 2\ f\ =\ -\ 8\ \hspace{3cm}\ c\ =\ 3$
$g\ =\ 0\ \hspace{3cm}\ f\ =\ -\ 4\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{(g^2\ +\ f^2\ -\ c)}$
$centre\ =\ (0,\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{(0^2\ +\ (-4)^2\ -\ 3)}$
$\hspace{6cm}\ r\ =\ \sqrt{(0\ +\ 16\ -\ 3)}$
$\hspace{6cm}\ r\ =\ \sqrt{13}$
$centre = (0, 4)\ \hspace{5cm}\ r\ =\ \sqrt{13}$

$\color {purple} {Example\ 6:}\ \color {red} {Show\ that\ the\ straight\ line}\ 4\ x-\ y\ =\ 17\ passes\ through\ \hspace{15cm}$$the\ centre\ of\ the\ circle\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ -\ 8\ \hspace{3cm}\ 2\ f\ =\ 2\ \hspace{3cm}\ c\ =\ 3$
$g\ =\ -\ 4\ \hspace{3cm}\ f\ =\ 1\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)$
$centre\ =\ (4,\ -\ 1)$
$4x\ -\ y=\ 17\ ——–(1)\ \hspace{10cm}$
$put\ x\ =\ 4,\ y\ =\ -\ 1\ in\ equation\ (1)$
$4(4)\ – (-1)\ =\ 17$
$17\ =\ 17$
$(4, -1)\ satisfies\ (1)$
$Hence\ the\ straight\ line\ 4\ x-\ y\ =\ 17\ passes\ through\ \hspace{15cm}$$the\ centre\ of\ the\ circle\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0$

$\color {purple} {Example\ 7:}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ \hspace{10cm}$$passing\ through\ the\ point\ A(2 ,-3)\ and\ having\ its\ centre\ at\ C(-5, 1)\ \hspace{5cm}$
$\color {blue} {Soln:}\ r\ =\ \sqrt{(-\ 5\ -\ 2)^2\ +\ (1\ +\ 3)^2}\ \hspace{15cm}$
$=\ \sqrt{(-7)^2\ +\ (4)^2}\ \hspace{10cm}$
$=\ \sqrt{49\ +\ 16}\ \hspace{10cm}$
$r=\ \sqrt{65}\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ -\ 5,\ k\ =\ 1\ (Given)\ and\ r\ =\ \sqrt{65}\ \hspace{10cm}$
$(x\ +\ 5)^2\ +\ (y\ -\ 1)^2\ =\ (\sqrt{65})^2\ \hspace{10cm}$
$x^2\ +\ 10\ x\ +\ 25\ +\ y^2\ -\ 2y\ +\ 1\ =\ 65\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 10\ x\ -\ 2y\ +\ 26\ -\ 65\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 10\ x\ -\ 2y\ -\ 39\ =\ 0\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 10\ x\ -\ 2y\ -\ 39\ =\ 0}\ \hspace{5cm}$

$\color {purple} {Example\ 8:}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ \hspace{10cm}$$having\ centre\ (2,-1)\ and\ passing\ through\ the\ point\ (8,7)\ \hspace{5cm}$
$\color {blue} {Soln:}\ r\ =\ \sqrt{(8\ -\ 2)^2\ +\ (7\ +\ 1)^2}\ \hspace{15cm}$
$=\ \sqrt{(6)^2\ +\ (8)^2}\ \hspace{10cm}$
$=\ \sqrt{36\ +\ 64}\ \hspace{10cm}$
$r=\ \sqrt{100}\ =\ 10\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ 2,\ k\ =\ -\ 1\ (Given)\ and\ r\ =\ 10\ \hspace{10cm}$
$(x\ -\ 2)^2\ +\ (y\ +\ 1)^2\ =\ 10^2\ \hspace{10cm}$
$x^2\ -\ 4\ x\ +\ 4\ +\ y^2\ +\ 2y\ +\ 1\ =\ 100\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 4\ x\ +\ 2y\ +\ 5\ -\ 100\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 4\ x\ +\ 2y\ -\ 95 =\ 0\ \hspace{10cm}$
$\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 4\ x\ +\ 2y\ -\ 95\ =\ 0}\ \hspace{5cm}$

#### Equation of circle on the line joining the points  ( x1,  y1 ) and ( x2,  y2 ) as diameter :

$(x\ -\ x_1)(x\ -\ x_2)\ +\ (y\ -\ y_1)(y\ -\ y_2)\ =\ 0$
$\color {purple} {Example\ 9:}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ on\ the\ line\ joining\ the\ points\ (2,3),\ (-\ 4,\ 5)\ as\ diameter.$$\color{red}{Aslo\ find\ the\ centre\ and\ radius}\ of\ the\ circle\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$(x\ -\ x_1)(x\ -\ x_2)\ +\ (y\ -\ y_1)(y\ -\ y_2)\ =\ 0$
$Given\ x_1\ =\ 2,\ y_1\ =\ 3,\ x_2\ =\ -\ 4,\ y_2\ =\ 5$
$(x\ -\ 2)(x\ +\ 4)\ +\ (y\ -\ 3)(y\ -\ 5)\ =\ 0$
$x^2\ +\ 4\ x\ – 2\ x\ – \ 8\ +\ y^2\ -\ 5y\ -\ 3y\ +\ 15\ =\ 0$
$x^2\ +\ 2\ x\ +\ y^2\ -\ 8y\ +\ 7\ =\ 0$
$x^2\ +\ y^2\ +\ 2\ x\ -\ 8y\ +\ 7\ =\ 0$
$\therefore\ the\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 2\ x\ -\ 8\ y\ +\ 7\ =\ 0}\ \hspace{5cm}$
$comparing\ with\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0$
$2\ g\ =\ 2\ \hspace{3cm}\ 2\ f\ =\ -\ 8\ \hspace{3cm}\ c\ =\ 7$
$g\ =\ 1\ \hspace{3cm}\ f\ =\ -\ 4\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}$
$centre\ =\ (-\ 1,\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{1^2\ +\ (-4)^2\ -\ 7}$
$\hspace{6cm}\ r\ =\ \sqrt{1\ +\ 16\ -\ 7}$
$\hspace{6cm}\ r\ =\ \sqrt{17\ -\ 7}$
$\hspace{6cm}\ r\ =\ \sqrt{10}$
$\boxed{centre = (- 1, 4)\ \hspace{4cm}\ r\ =\ \sqrt{10}}$

$\color {purple} {Example\ 10}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ two\ of\ whose\ diameters\ are\ \hspace{15cm}$$x\ +\ y\ =\ 6\ and\ x\ +\ 2y=\ 4\ and\ its\ radius\ is\ 10\ units\ \hspace{10cm}$
$\color {blue} {Soln:}\ x\ +\ y\ =\ 6\ ————–\ (1)\ \hspace{15cm}$
$x\ +\ 2y\ =\ 4\ ————–\ (2)\ \hspace{15cm}$
$(1)\ -\ (2)\ \implies\ -\ y\ =\ 2\ \hspace{10cm}$
$\therefore\ \boxed{y\ =\ -\ 2}\ \hspace{10cm}$
$Substitute\ y\ =\ -\ 2\ in\ (1)\ \hspace{10cm}$
$x\ -\ 2\ =\ 6\ \hspace{10cm}$
$x\ =\ 6\ +\ 2\ \hspace{10cm}$
$\therefore\ \boxed{x\ =\ 8}\ \hspace{10cm}$
$Centre\ =\ (8,\ -2)\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ 8,\ k\ =\ -\ 2\ and\ r\ =\ 10\ \hspace{10cm}$
$(x\ -\ 8)^2\ +\ (y\ +\ 2)^2\ =\ 10^2\ \hspace{10cm}$
$x^2\ -\ 16\ x\ +\ 64\ +\ y^2\ -\ 4y\ +\ 4\ =\ 100\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 16\ x\ -\ 4y\ +\ 68\ =\ 100\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 16\ x\ -\ 10y\ +\ 68\ -\ 100\ =\ 0\ \hspace{10cm}$
$\therefore\ The\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 16\ x\ -\ 10\ y\ -\ 32\ =\ 0}\ \hspace{5cm}$

FAMILY OF CIRCLES

Concentric Circles:

Two or more circles having the same centre but differ in radii are called concentric circles.

Note:   Equation of concentric circles differ only by the constant term.

$\color {purple} {Example\ 11\ .}\ \color {red} {Show\ that\ the\ circles}\ x^2\ +\ y^2\ -\ 4\ x\ +\ 2\ y\ +\ 5\ =\ 0\ \hspace{5cm}$$and\ x^2\ +\ y^2\ -\ 4\ x\ +\ 2\ y\ +\ 5\ =\ 0\ are\ concentric\ circles\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{19cm}$

From the two given equations of the circles, we observe that the constant term alone differs

∴The given circles are concentric circles.

$\color {purple} {Example\ 12\ .}\ \color {red} { Find\ the\ equation\ of\ the\ circle}\ passing\ through\ the\ point\ (5 ,4)\ \hspace{7cm}$$and\ concentric\ to\ the\ circle\ x^2\ +\ y^2\ -\ 8\ x\ +\ 12\ y\ +\ 15\ =\ 0\ \hspace{5cm}$
$\color {blue} {Soln:}\ Equation\ of\ concentric\ circle\ be\ x^2\ +\ y^2\ -\ 8\ x\ +\ 12\ y\ +\ k\ =\ 0\ ———- (1)\ \hspace{6cm}$
$Put\ x\ = 5,\ y\ =\ 4\ in\ equation\ ( 1 )\ \hspace{10cm}$
$(5)^2\ +\ (4)^2\ -\ 8(5)\ +\ 12(4)\ +\ K\ =\ 0\ \hspace{10cm}$
$25\ +\ 16\ -\ 40\ +\ 48\ +\ K\ =\ 0\ \hspace{10cm}$
$49\ +\ K\ =\ 0\ \hspace{10cm}$
$K\ =\ -\ 49\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 8\ x\ +\ 12\ y\ -\ 49\ =\ 0}\ \hspace{5cm}$

Orthogonal Circles:

Two circles   x2  +   y2  + 2g1x  +2f1y+c1 = 0 and  x2  +   y2  + 2g2x  + 2f2y +  c2 = 0 are said to be Orthogonal Circles if

2g1g2  + 2 f1 f2  =  c1 + c2

$\color {purple} {Example\ 13\ .}\ \color {red} {Prove\ that\ the\ circles}\ x^2\ +\ y^2\ -\ 4\ x\ +\ 6\ y\ +\ 4\ =\ 0\ \hspace{7cm}$$and\ x^2\ +\ y^2\ +\ 2\ x\ +\ 4\ y\ +\ 4\ =\ 0\ \color {red} {cut\ orthogonally}\ \hspace{8cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$Given\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 4\ x\ +\ 6\ y\ +\ 4\ =\ 0 ——— (1)$
$x^2\ +\ y^2\ +\ 2\ x\ +\ 4\ y\ +\ 4\ =\ 0——— (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 =\ -\ 4\ \hspace 2cm\ 2f_1\ =\ 6\ \hspace 2cm\ c_1 =\ 4$
$g_1 =\ -\ 2\ \hspace 2cm\ f_1 =\ 3\ \hspace 2cm\ c_1 =\ 4$
$From\ (2)\ \hspace 10cm$
$2g_2 =\ 2\ \hspace 2cm\ 2f_2 =\ 4\ \hspace 2cm\ c_2 =\ 4$
$g_2 =\ 1\ \hspace 2cm\ f_2 =\ 2\ \hspace 2cm\ c_2 =\ 4$
$The\ condition\ for\ orthogonally\ is$
$2\ g_1\ g_2\ +\ 2\ f_1\ f_2\ =\ c_1\ +\ c_2$
$2\ (-2)(1)\ +\ 2\ (3) (2)\ =\ 4\ +\ 4$
$-\ 4\ +\ 12 =\ 8$
$8\ =\ 8$
$\therefore\ The\ given\ two\ circles\ cut\ orthogonally$

$\color {purple} {Example\ 14\ .}\ \color{red}{Find\ the\ equation\ of\ the\ circle}\ passing\ through\ the\ origin\ and\ cuts\ orthogonally\ \hspace{3cm}$$each\ of\ the\ circles\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0\ and\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0$
$\color {blue} {Soln:}\ \hspace{20cm}$
$Let\ the\ equation\ of\ circle\ be\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ ——-(1)$
$(1)\ passes through (0,\ 0)$
$(0)^2\ +\ (0)^2\ +\ 2\ g(0)\ +\ 2f(0)\ +\ c\ =\ 0$
$c\ =\ 0$
$Given\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0 ——— (2)$
$x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0——— (3)$
$From\ (2)\ \hspace 10cm$
$2g_1 = -\ 6\ \hspace 2cm\ 2f_1 = 0\ \hspace 2cm\ c_1 = 8$
$g_1 =\ -\ 3\ \hspace 2cm\ f_1 = 0\ \hspace 2cm\ c_1 =\ 8$
$From\ (3)\ \hspace 10cm$
$2g_2 = -2\ \hspace 2cm\ 2f_2 = -2\ \hspace 2cm\ c_2 = – 7$
$g_2 = -1\ \hspace 2cm\ f_2 = -1\ \hspace 2cm\ c_2 = – 7$
$Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0$
$2\ g\ g_1\ +\ 2\ f\ f_1\ =\ c\ +\ c_1$
$2\ g(-\ 3)\ +\ 2\ f (0)\ =\ c\ +\ 8$
$-\ 6g\ +\ 0\ =\ 0\ +\ 8$
$-\ 6g\ =\ 8$
$g\ =\ -\ \frac{8}{6}$
$\boxed{g\ =\ -\ \frac{4}{3}}$
$Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0$
$2\ g\ g_2\ +\ 2\ f\ f_2\ =\ c\ +\ c_2$
$2\ g(-\ 1)\ +\ 2\ f (- 1)\ =\ c\ -\ 7$
$-\ 2g\ -\ 2\ f\ =\ 0\ -\ 7$
$-\ 2(-\ \frac{4}{3})\ -\ 2\ f\ =\ -\ 7$
$\frac{8}{3}\ -\ 2\ f\ =\ -\ 7$
$2\ f\ =\ \frac{8}{3}\ +\ 7$
$2\ f\ =\ \frac{8\ +\ 21}{3}$
$2f\ =\ \frac{29}{3}$
$\boxed{f\ =\ \frac{29}{6}}$
$Required\ equation\ of\ the\ circle\ is$
$x^2\ +\ y^2\ +\ 2\ (\frac{-4}{3})\ x\ +\ 2\ (\frac{29}{6})\ y\ +\ 0\ =\ 0$
$x^2\ +\ y^2\ -\ \frac{8}{3}\ x\ +\ \frac{29}{3}\ y\ =\ 0$
$3x^2\ +\ 3y^2\ -\ 8\ x\ +\ 29\ y\ =\ 0$

### Contact of Circles:

Case ( i ) :

Two circles touch externally if the distance between their centres is equal to sum of their radii.

i.e  C1C=  r1  +  r2

Co-ordinates of point of contact are

P = ( (r1 x2 + r2 x1 ) / (r1  +  r2 ) ,  (r1 y2 + r2 y1 ) / (r1  +  r2 ) )

where  C1 = (x1,  y1)   and   C2  =  (x2,  y2)

Case ( ii ) :

Two circles touch internally if the distance between their centres is equal to difference of their radii.

i.e  C1C= r1  –  ror   r2  –  r1

Co-ordinates of point of contact are

P = ( (r1 x2 – r2 x1 ) / (r1  –  r2 ) ,  (r1 y2 – r2 y1 ) / (r1  –  r2 ) )

where  C1 = (x1,  y1)   and   C2  =  (x2,  y2)

$\color {purple} {Example\ 15\ .}\ \color {red} {Prove\ that}\ the\ circles\ x^2\ +\ y^2\ + 2x\ -\ 4y\ -\ 3\ = 0\ \hspace{5cm}$$and\ x^2\ +\ y^2\ -\ 8x\ +\ 6y\ +\ 7\ = 0\ touch\ each\ other.\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace {19cm}$
$Given\ x^2 + y^2 + 2x\ -\ 4y\ -\ 3\ = 0 —–(1)$
$Given\ x^2\ +\ y^2\ -\ 8x\ +\ 6y\ +\ 7\ = 0 — (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 = 2\ \hspace 2cm\ 2f_1 =\ -\ 4\ \hspace 2cm\ c_1 = -\ 3$
$g_1 = 1\ \hspace 2cm\ f_1 = -\ 2\ \hspace 2cm\ c_1 = – 3$
$Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}$
$C_1 = (- 1,\ 2)\ \hspace 10cm\ r_1 = \sqrt{(1)^2 + (-2)^2 + 3}$
$\hspace 10cm\ r_1 = \sqrt{1\ +\ 4\ +\ 3}$
$\hspace 10cm\ r_1 = \sqrt{8}\ =\ 2\ \sqrt{2}$
$\boxed{C_1 = ( -1, 2)\ and\ r_1 =\ 2\ \sqrt{2}}$
$From\ (2)\ \hspace 10cm$
$2g_2 = -\ 8\ \hspace 2cm\ 2f_2 = 6\ \hspace 2cm\ c_2 =\ 7$
$g_2 = -\ 4\ \hspace 2cm\ f_2 =\ 3\ \hspace 2cm\ c_2 =\ 7$
$Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}$
$C_2 =\ (4,\ -3)\ \hspace 10cm\ r_2 = \sqrt{(-4)^2 +\ (3)^2\ -\ 7}$
$\hspace 10cm\ r_2 = \sqrt{16\ +\ 9\ -\ 7}$
$\hspace 10cm\ r_2 = \sqrt{18} =\ 3\ \sqrt{2}$
$\boxed{C_2 = ( 3, -3)\ and\ r_2 = 8}$
$C_1C_2 = \sqrt{(-1-4)^2 +\ (2\ +\ (3))^2}$
$C_1C_2 = \sqrt{(-5)^2 + (5)^2}$
$C_1C_2 = \sqrt{25\ +\ 25}$
$C_1C_2 = \sqrt{50} =\ 5\ \sqrt{2}$
$r_1\ +\ r_2\ =\ 3\ \sqrt{2}\ +\ 2\ \sqrt{2}\ =\ 5\ \sqrt{2}\ =\ C_1C_2$
$\boxed{C_1C_2\ =\ r_1\ +\ r_2}$
$The\ given\ circles\ touch\ each\ other\ externally$

$\color {purple} {Example\ 16\ }\ \color {red} {Prove\ that}\ the\ circles\ x^2\ +\ y^2\ -\ 4x\ +\ 6y\ -\ 112\ = 0\ \hspace{5cm}$$and\ x^2\ +\ y^2\ -\ 10x\ -\ 6y\ +\ 14\ = 0\ touch\ each\ other.\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace {19cm}$
$Given\ x^2 + y^2 -\ 4x\ +\ 6y\ -\ 112\ = 0 ——————— (1)$
$Given\ x^2 + y^2 – 4x\ +\ 6y\ -\ 112\ = 0 ——————— (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 =\ -\ 4\ \hspace 2cm\ 2f_1\ =\ 6\ \hspace 2cm\ c_1 =\ -\ 112$
$g_1 =\ -\ 2\ \hspace 2cm\ f_1 =\ 3\ \hspace 2cm\ c_1 =\ -\ 112$
$Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}$
$C_1 = (2,\ -\ 3)\ \hspace 10cm\ r_1 = \sqrt{(-2)^2\ +\ (3)^2\ +\ 112}$
$\hspace 10cm\ r_1 = \sqrt{4\ +\ 9\ +\ 112}$
$\hspace 10cm\ r_1 = \sqrt{125} =\ \sqrt{25\ ×\ 5}\ =\ 5\ \sqrt{5}$
$\boxed{C_1 = ( 2, -3)\ and\ r_1 =\ 5\ \sqrt{5}}$
$From\ (2)\ \hspace 10cm$
$2g_2 =\ -\ 10\ \hspace 2cm\ 2f_2\ =\ -\ 6\ \hspace 2cm\ c_2 =\ 14$
$g_2 =\ -\ 5\ \hspace 2cm\ f_2 =\ -\ 3\ \hspace 2cm\ c_2 =\ 14$
$Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}$
$C_2 = (5\, 3)\ \hspace 10cm\ r_2 = \sqrt{(-5)^2\ +\ (-3)^2\ -\ 14}$
$\hspace 10cm\ r_2\ = \sqrt{25\ + 9\ -\ 14}$
$\hspace 10cm\ r_2 = \sqrt{20} =\ \sqrt{4\ ×\ 5}\ =\ 2\ \sqrt{5}$
$\boxed{C_2 = ( 5, 3)\ and\ r_2 =\ 2\ \sqrt{5}}$
$C_1C_2 = \sqrt{(5\ -\ 2)^2\ +\ (3\ +\ (3))^2}$
$C_1C_2 = \sqrt{(3)^2\ +\ (6)^2}$
$C_1C_2 = \sqrt{9\ +\ 36}$
$C_1C_2 = \sqrt{45}\ =\ \sqrt{9\ ×\ 5}\ =\ 3\ \sqrt{5}$
$r_1 – r_2\ =\ 5\ \sqrt{5}\ -\ 2\ \sqrt{5} =\ 3\ \sqrt{5}\ =\ C_1C_2$
$\boxed{C_1C_2 = r_1 – r_2}$
$The\ given\ circles\ touch\ each\ other\ internally$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ with\ centre\ (1, -2)\ and\ radius\ 5\ units.\ \hspace{5cm}$
$\color {purple} {2\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ with\ centre\ (0, -3)\ and\ radius\ 2\ units.\ \hspace{5cm}$
$\color {purple} {3\ .}\ \color {red} {Write\ down\ the\ formula\ for\ center\ and\ radius\ of\ the\ circle}\ \hspace{7cm}$$x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$\color {purple} {4\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ \hspace{10cm}$$x^2\ +\ y^2\ +\ 10\ x\ +\ 8\ y\ +\ 5\ =\ 0\ \hspace{5cm}$
$\color {purple} {5\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ x^2\ +\ y^2\ +\ 4\ x\ +\ 4\ y\ -\ 1\ =\ 0\ \hspace{5cm}$
$\color {purple} {6\ .}\ \color {red} {Show\ that}\ the\ circles\ x^2\ +\ y^2\ -\ 2\ x\ +\ 4\ y\ -\ 3\ =\ 0\ and\ \hspace{7cm}$$x^2\ +\ y^2\ -\ 2\ x\ +\ 4\ y\ +\ 5\ =\ 0\ \color {red} {are\ concentric\ circles}\ \hspace{5cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {7\ .}\ \color {red} {Show\ that}\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ \color {red} {is\ a\ diameter\ of\ the\ circle}\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{5cm}$
$\color {purple} {8\ .}\ \color {red} { Find\ the\ equation\ of\ the\ circle}\ passing\ through\ the\ point\ (1 ,1)\ \hspace{7cm}$$and\ concentric\ to\ the\ circle\ x^2\ +\ y^2\ +\ 4\ x\ +\ 6\ y\ -\ 15\ =\ 0\ \hspace{5cm}$
$\color {purple} {9\ .}\ \color {red} { Find\ the\ equation\ of\ the\ circle}\ concentric\ with\ the\ circle\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{7cm}$$and\ passing\ through\ the\ point\ (1,1)\ \hspace{5cm}$
$\color {purple} {10\ .}\ \color {red} { Find\ the\ equation\ of\ the\ circle}\ concentric\ with\ the\ circle\ x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ -\ 9\ =\ 0\ \hspace{7cm}$$and\ passing\ through\ the\ point\ (-4 ,-5)\ \hspace{5cm}$
$\color {purple} {11\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ concentric\ with\ the\ circle\ x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ -\ 23\ =\ 0\ \hspace{7cm}$$and\ having\ radius\ 3\ units\ \hspace{5cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {12\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ passing\ through\ the\ point\ (-7.1)\ \hspace{7cm}$$and\ having\ its\ centre\ at\ (-4,-3)\ \hspace{5cm}$
$\color {purple} {13\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ two\ of\ whose\ diameters\ are\ \hspace{15cm}$$2x\ -\ 3y\ +\ 1\ =\ 0\ and\ x\ +\ 2y\ -\ 17\ =\ 0\ and\ its\ radius\ is\ 8\ units\ \hspace{10cm}$
$\color {purple} {14\ .}\ \color{red}{Find\ the\ equation\ of\ the\ circle}\ passing\ through\ the\ origin\ and\ cuts\ orthogonally\ \hspace{3cm}$$each\ of\ the\ circles\ x^2\ +\ y^2\ -\ 8\ y\ +\ 12\ =\ 0\ and\ x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ -\ 3\ =\ 0$
$\color {purple} {15\ .}\ \color {red} {Show\ that}\ the\ circles\ x^2 + y^2 + 2x – 8 = 0\ and\ x^2 + y^2 – 6x + 6y -46 = 0\ touch\ each\ other.\ \hspace10cm$
$\color {purple} {16\ .}\ \color {red} {Prove\ that}\ the\ circles\ x^2 + y^2\ -\ 2x\ +\ 6y\ +\ 6 = 0\ and\ \hspace {8cm}$$x^2 + y^2\ -\ 5x + 6y\ +\ 15\ = 0\ touch\ each\ other\ \hspace {7cm}$
$\color {purple} {17\ .}\ \color {red} {Prove\ that\ the\ circles}\ x^2\ +\ y^2\ -\ 10\ x\ -\ 24\ y\ +\ 120\ =\ 0\ and\ \hspace{7cm}$$x^2\ +\ y^2\ =\ 400\ \color {red} {touch\ each\ other}\ \hspace{5cm}$