# 1.1 – N – Analytical Geometry – II Exercise Problems With Solutions

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ with\ centre\ (1, -2)\ and\ radius\ 5\ units.\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ centre\ =\ (1, -2)\ \hspace{4cm}\ and\ radius\ =\ 5\ \hspace{6cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2$
$Given\ h\ =\ 1,\ k = – 2\ and\ r\ =\ 5$
$(x\ -\ 1)^2\ +\ (y\ +\ 2)^2\ =\ 5^2$
$x^2\ -\ 2\ x\ +\ 1\ +\ y^2\ +\ 4y\ +\ 4\ =\ 25\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 2\ x\ +\ 4y\ +\ 1\ +\ 4\ -\ 25\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 2\ x\ +\ 4y\ -\ 20\ =\ 0\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 2\ x\ +\ 4\ y\ -\ 20\ =\ 0}\ \hspace{5cm}$

$\color {purple} {2\ .}\ \color {red} {Find\ the\ equation\ of\ the\ circle}\ with\ centre\ (0, -3)\ and\ radius\ 2\ units.\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ centre\ =\ (0, -3)\ \hspace{4cm}\ and\ radius\ =\ 2\ \hspace{6cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2$
$Given\ h\ =\ 0,\ k = – 3\ and\ r\ =\ 2$
$(x\ -\ 0)^2\ +\ (y\ +\ 3)^2\ =\ 2^2$
$x^2\ +\ y^2\ +\ 6y\ +\ 9\ =\ 4\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 6y\ +\ 9\ -\ 4\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 6y\ +\ 5\ =\ 0\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 6\ y\ +\ 5\ =\ 0}\ \hspace{5cm}$

$\color {purple} {3\ .}\ \color {red} {Write\ down\ the\ formula\ for\ center\ and\ radius\ of\ the\ circle}\ \hspace{7cm}$$x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$\color {blue} {Soln:}\ centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}\ \hspace{6cm}$
$\color {purple} {4\ .}\ \color {red} {Find\ the\ centre\ and\ radius\ of\ the\ circle}\ \hspace{10cm}$$x^2\ +\ y^2\ +\ 10\ x\ +\ 8\ y\ +\ 5\ =\ 0\ \hspace{5cm}$
$\color {blue} { Soln:}\ Given\ x^2\ +\ y^2\ +\ 10\ x\ +\ 8\ y\ +\ 5\ =\ 0\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ 10\ \hspace{3cm}\ 2\ f\ =\ 8\ \hspace{3cm}\ c\ =\ 5$
$g\ =\ 5\ \hspace{3cm}\ f\ =\ 4\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{(g^2\ +\ f^2\ -\ c)}$
$centre\ =\ (-\ 5,\ -\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{(5^2\ +\ 4^2\ -\ 5)}$
$\hspace{6cm}\ r\ =\ \sqrt{(25\ +\ 16\ -\ 5)}$
$\hspace{6cm}\ r\ =\ \sqrt{36}\ =\ 6$

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

$\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}$
$\color {blue} {Soln:}\ \hspace{19cm}$

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

∴The given circles are concentric circles.

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {7\ .}\ \color {red} {Show\ that}\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ \color {red} {is\ a\ diameter\ of\ the\ circle}\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{5cm}$
$\color {blue} {Soln:}\ Given\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{10cm}$
$W.\ K.\ T.\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}$
$2\ g\ =\ -\ 6\ \hspace{6cm}\ 2\ f\ =\ 10\ \hspace{3cm}$
$g\ =\ -\ 3\ \hspace{3cm}\ f\ =\ 5\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{8cm}$
$=\ (3,\ -\ 5)\ \hspace{8cm}$
$Put\ x\ =\ 3,\ y\ =\ -\ 5\ in\ \hspace{8cm}$
$2\ (3)\ +\ 3\ (-5)\ +\ 9\ =\ 0\ \hspace{8cm}$
$6\ -\ 15\ +\ 9\ =\ 0\ \hspace{8cm}$
$-\ 9\ +\ 9\ =\ 0\ \hspace{8cm}$
$0\ =\ 0\ \hspace{8cm}$
$\therefore\ (3, -\ 5)\ lies\ on\ the\ line\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ \hspace{5cm}$
$\therefore\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ is\ the\ diameter\ of\ the\ circle\ \hspace{5cm}$

$\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}$
$\color {blue} {Soln:}\ Equation\ of\ concentric\ circle\ be\ x^2\ +\ y^2\ +\ 4\ x\ +\ 6\ y\ +\ k\ =\ 0\ — (1)\ \hspace{6cm}$
$Put\ x\ = 1,\ y\ =\ 1\ in\ equation\ ( 1 )\ \hspace{10cm}$
$(1)^2\ +\ (1)^2\ +\ 4(1)\ +\ 6(1)\ +\ K\ =\ 0\ \hspace{10cm}$
$1\ +\ 1\ +\ 4\ +\ 6\ +\ K\ =\ 0\ \hspace{10cm}$
$12\ +\ K\ =\ 0\ \hspace{10cm}$
$K\ =\ -\ 12\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$

The required equation of the circle is  x2  +   y2   +  4x  +  6y   – 12 = 0

$\boxed{x^2\ +\ y^2\ +\ 4\ x\ +\ 6\ y\ -\ 12\ =\ 0}\ \hspace{5cm}$

$\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}$
$\color {blue} {Soln:}\ Equation\ of\ concentric\ circle\ be\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ +\ k\ =\ 0\ ———- (1)\ \hspace{6cm}$
$Put\ x\ = 1,\ y\ =\ 1\ in\ equation\ ( 1 )\ \hspace{10cm}$
$(1)^2\ +\ (1)^2\ -\ 6(1)\ +\ 10(1)\ +\ K\ =\ 0\ \hspace{10cm}$
$1\ +\ 1\ -\ 6\ +\ 10\ +\ K\ =\ 0\ \hspace{10cm}$
$6\ +\ K\ =\ 0\ \hspace{10cm}$
$K\ =\ -\ 6\ \hspace{10cm}$
$\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 6\ =\ 0}\ \hspace{5cm}$

$\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}$
$\color {blue} {Soln:}\ Equation\ of\ concentric\ circle\ be\ x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ +\ k\ =\ 0\ ———- (1)\ \hspace{6cm}$
$Put\ x\ = – 4,\ y\ =\ – 5\ in\ equation\ ( 1 )\ \hspace{10cm}$
$(- 4)^2\ +\ (- 5)^2\ -\ 4(- 4)\ -\ 6(-5)\ +\ K\ =\ 0\ \hspace{10cm}$
$16\ +\ 25\ +\ 16\ +\ 30\ +\ K\ =\ 0\ \hspace{10cm}$
$87\ +\ K\ =\ 0\ \hspace{10cm}$
$K\ =\ -\ 87\ \hspace{10cm}$
$\therefore\ The\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ -\ 87\ =\ 0}\ \hspace{5cm}$

$\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}$$and\ having\ radius\ 3\ units\ \hspace{5cm}$
$\color {blue} {Soln:}\ As\ we\ know\ that\ centres\ of concentric\ circles\ is\ same, \hspace{15cm}$$we\ will\ find\ the\ centre\ for\ x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ -\ 23\ =\ 0\ \hspace{5cm}$
$W.\ K.\ T.\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$2\ g\ =\ 8\ \hspace{6cm}\ 2\ f\ =\ -\ 4\ \hspace{3cm}$
$g\ =\ 4\ \hspace{3cm}\ f\ =\ -\ 2\ \hspace{3cm}$
$centre\ =\ (-\ g,\ -\ f)\ \hspace{8cm}$
$=\ (-\ 4,\ 2)\ \hspace{8cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ -\ 4,\ k\ =\ 2\ and\ r\ =\ 3\ (Given)\ \hspace{10cm}$
$(x\ +\ 4)^2\ +\ (y\ -\ 2)^2\ =\ 3^2\ \hspace{10cm}$
$x^2\ +\ 8\ x\ +\ 16\ +\ y^2\ -\ 4y\ +\ 4\ =\ 9\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 8\ x\ -\ 4y\ +\ 16\ +\ 4\ -\ 9\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 8\ x\ -\ 4y\ +\ 11\ =\ 0\ \hspace{10cm}$
$\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ +\ 11\ =\ 0}\ \hspace{5cm}$

$\LARGE{\color {purple} {PART- C}}$
$\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}$
$\color {blue} {Soln:}\ r\ =\ \sqrt{(-\ 4\ +\ 7)^2\ +\ (-3\ -\ 1)^2}\ \hspace{15cm}$
$=\ \sqrt{(3)^2\ +\ (-\ 4)^2}\ \hspace{10cm}$
$=\ \sqrt{9\ +\ 16}\ \hspace{10cm}$
$r=\ \sqrt{25}\ =\ 5\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ -\ 4,\ k\ =\ -\ 3\ (Given)\ and\ r\ =\ 5\ \hspace{10cm}$
$(x\ +\ 4)^2\ +\ (y\ +\ 3)^2\ =\ 5^2\ \hspace{10cm}$
$x^2\ +\ 8\ x\ +\ 16\ +\ y^2\ +\ 6y\ +\ 9\ =\ 25\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 8\ x\ +\ 6y\ +\ 25\ -\ 25\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ +\ 8\ x\ +\ 6y\ =\ 0\ \hspace{10cm}$
$\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ +\ 8\ x\ +\ 6\ y\ =\ 0}\ \hspace{5cm}$

$\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}$
$\color {blue} {Soln:}\ 2x\ -\ 3y\ =\ -\ 1\ ————–\ (1)\ \hspace{15cm}$
$x\ +\ 2y\ =\ 17\ ————–\ (2)\ \hspace{15cm}$
$(2) \times\ 2\ and\ (1)\ -\ (2)\ \implies\ -\ 7y\ =\ -\ 35\ \hspace{10cm}$
$\therefore\ \boxed{y\ =\ 5}\ \hspace{10cm}$
$Substitute\ y\ =\ 5\ in\ (1)\ \hspace{10cm}$
$2\ x\ -\ 15\ =\ -1\ \hspace{10cm}$
$2\ x\ =\ 14\ \hspace{10cm}$
$\therefore\ \boxed{x\ =\ 7}\ \hspace{10cm}$
$Centre\ =\ (7,\ 5)\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ 7,\ k\ =\ 5\ and\ r\ =\ 8\ \hspace{10cm}$
$(x\ -\ 7)^2\ +\ (y\ -\ 5)^2\ =\ 8^2\ \hspace{10cm}$
$x^2\ -\ 14\ x\ +\ 49\ +\ y^2\ -\ 10y\ +\ 25\ =\ 64\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 14\ x\ -\ 10y\ +\ 74\ -\ 64\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 14\ x\ -\ 10y\ +\ 10 =\ 0\ \hspace{10cm}$
$\therefore\ The\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 14\ x\ -\ 10\ y\ +\ 10\ =\ 0}\ \hspace{5cm}$

$\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$
$\color {blue} {Soln:}\ \hspace{20cm}$
$Let\ the\ equation\ of\ circle\ be\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ ——-(1)$
$(1)\ passes through (0,\ 0)$
$(0)^2\ +\ (0)^2\ +\ 2\ g(0)\ +\ 2f(0)\ +\ c\ =\ 0$
$c\ =\ 0$
$Given\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 8\ y\ +\ 12\ =\ 0 ——— (2)$
$x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ -\ 3\ =\ 0——— (3)$
$From\ (2)\ \hspace 10cm$
$2g_1 =\ 0\ \hspace 2cm\ 2f_1 = -\ 8\ \hspace 2cm\ c_1 = 12$
$g_1 =\ 0\ \hspace 2cm\ f_1 = -\ 4\ \hspace 2cm\ c_1 =\ 12$
$From\ (3)\ \hspace 10cm$
$2g_2 = -4\ \hspace 2cm\ 2f_2 = -6\ \hspace 2cm\ c_2 = – 3$
$g_2 = -2\ \hspace 2cm\ f_2 = -3\ \hspace 2cm\ c_2 =\ -\ 3$
$Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 8\ y\ +\ 12\ =\ 0$
$2\ g\ g_1\ +\ 2\ f\ f_1\ =\ c\ +\ c_1$
$2\ g(0)\ +\ 2\ f (-4)\ =\ c\ +\ 12$
$0\ -\ 8\ f\ =\ 12$
$-\ 8f\ =\ 12$
$f\ =\ -\ \frac{12}{8}$
$\boxed{f\ =\ -\ \frac{3}{2}}$
$Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 4\ x\ -\ 6\ y\ -\ 3\ =\ 0$
$2\ g\ g_2\ +\ 2\ f\ f_2\ =\ c\ +\ c_2$
$2\ g(-\ 2)\ +\ 2\ f (- 3)\ =\ 0\ -\ 3$
$-\ 4g\ -\ 6\ f\ =\ -\ 3$
$-\ 4g\ -\ 6(\ -\ \frac{3}{2})\ =\ 3$
$-\ 4g\ +\ 9\ =\ -\ 3$
$-\ 4g\ =\ -\ 3\ -\ 9$
$-\ 4g\ =\ -\ 12$
$g\ =\ \frac{-12}{-4}$
$\boxed{g\ =\ 3}$
$Required\ equation\ of\ the\ circle\ is$
$x^2\ +\ y^2\ +\ 2\ (3)\ x\ +\ 2\ (\frac{-3}{2}))\ y\ +\ 0\ =\ 0$
$x^2\ +\ y^2\ +\ 6\ x\ -\ 3\ y\ =\ 0$

$\color {purple} {15\ .}\ \color {red} {Show\ that}\ the\ circles\ x^2 + y^2 + 2x – 8 = 0\ \hspace{8cm}$$and\ x^2 + y^2 – 6x + 6y -46 = 0\ touch\ each\ other.\ \hspace{7cm}$
$\color {blue} {Soln:}\ \hspace {19cm}$
$Given\ x^2 + y^2 + 2x – 8 = 0 ——————— (1)$
$Given\ x^2 + y^2 – 6x + 6y -46 = 0 ——————— (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 = 2\ \hspace 2cm\ 2f_1 = 0\ \hspace 2cm\ c_1 = – 8$
$g_1 = 1\ \hspace 2cm\ f_1 = 0\ \hspace 2cm\ c_1 = – 8$
$Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}$
$C_1 = (- 1,\ 0)\ \hspace 10cm\ r_1 = \sqrt{1^2 + 0^2 + 8}$
$\hspace 10cm\ r_1 = \sqrt{1 + 0 + 8}$
$\hspace 10cm\ r_1 = \sqrt{9} = 3$
$\boxed{C_1 = ( -1, 0)\ and\ r_1 = 3}$
$From\ (2)\ \hspace 10cm$
$2g_2 = -6\ \hspace 2cm\ 2f_2 = 6\ \hspace 2cm\ c_2 = – 46$
$g_2 = -3\ \hspace 2cm\ f_2 = 3\ \hspace 2cm\ c_2 = – 46$
$Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}$
$C_2 = (3,\ -3)\ \hspace 10cm\ r_2 = \sqrt{(-3)^2 + 3^2 + 46}$
$\hspace 10cm\ r_2 = \sqrt{9 + 9 + 46}$
$\hspace 10cm\ r_2 = \sqrt{64} = 18$
$\boxed{C_2 = ( 3, -3)\ and\ r_2 = 8}$
$C_1C_2 = \sqrt{(-1-3)^2 + (0- (-3))^2}$
$C_1C_2 = \sqrt{(-4)^2 + (3)^2}$
$C_1C_2 = \sqrt{16 + 9}$
$C_1C_2 = \sqrt{25} = 5$
$r_2 – r_1 = 8 – 3 = 5 = C_1C_2$
$\boxed{C_1C_2 = r_2 – r_1}$
$The\ given\ circles\ touch\ each\ other\ internally$

$\color {purple} {16\ .}\ \color {red} {Prove\ that}\ the\ circles\ x^2 + y^2\ -\ 2x\ +\ 6y\ +\ 6 = 0\ and\ \hspace {5cm}$$x^2 + y^2\ -\ 5x + 6y\ +\ 15\ = 0\ touch\ each\ other\ \hspace {7cm}$
$\color {blue} {Soln:}\ \hspace {19cm}$
$Given\ x^2 + y^2 – 2x\ +\ 6y\ +\ 6\ = 0 —–(1)$
$Given\ x^2\ +\ y^2\ -\ 5x\ +\ 6y\ +\ 15\ = 0 — (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 =\ -\ 2\ \hspace 2cm\ 2f_1 =\ 6\ \hspace 2cm\ c_1 =\ 6$
$g_1 =\ -\ 1\ \hspace 2cm\ f_1 =\ 3\ \hspace 2cm\ c_1 =\ 6$
$Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}$
$C_1 = (1,\ -\ 3)\ \hspace 10cm\ r_1 = \sqrt{(-1)^2 + (3)^2\ -\ 6}$
$\hspace 10cm\ r_1 = \sqrt{1\ +\ 9\ -\ 6}$
$\hspace 10cm\ r_1 = \sqrt{4}\ =\ 2$
$\boxed{C_1 = (1, -\ 3)\ and\ r_1 =\ 2}$
$From\ (2)\ \hspace 10cm$
$2g_2 = -\ 5\ \hspace 2cm\ 2f_2 =\ 6\ \hspace 2cm\ c_2 =\ 15$
$g_2 = -\ \frac{5}{2}\ \hspace 2cm\ f_2 =\ 3\ \hspace 2cm\ c_2 =\ 15$
$Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}$
$C_2 =\ (\frac{5}{2},\ -3)\ \hspace 10cm\ r_2 = \sqrt{(-\ \frac{5}{2})^2 +\ (3)^2\ -\ 15}$
$\hspace 10cm\ r_2 = \sqrt{\frac{25}{4}\ +\ 9\ -\ 15}$
$\hspace 10cm\ r_2 = \sqrt{\frac{25}{4}\ -\ 6}$
$\hspace 10cm\ r_2 = \sqrt{\frac{25\ -\ 24}{4}}$
$\hspace 10cm\ r_2 = \sqrt{\frac{1}{4}}=\ \frac{1}{2}$
$\boxed{C_2 = (\frac{5}{2},\ -3)\ and\ r_2 =\ \frac{1}{2}}$
$C_1C_2 = \sqrt{(\frac{5}{2}\ -\ 1)^2 +\ (-\ 3\ +\ 3)^2}$
$C_1C_2 = \sqrt{(\frac{5\ -\ 2}{2})^2}$
$C_1C_2 = \sqrt{(\frac{3}{2})^2}$
$C_1C_2 =\ \frac{3}{2}$
$r_1\ -\ r_2\ =\ 2\ -\ \frac{1}{2}\ =\ \frac{4\ -\ 1}{2}\ =\ \frac{3}{2}\ =\ C_1C_2$
$\boxed{C_1C_2\ =\ r_1\ -\ r_2}$
$The\ given\ circles\ touch\ each\ other\ internally$

$\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}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$Given\ x^2\ +\ y^2\ -\ 10\ x\ -\ 24\ y\ +\ 120\ =\ 0 ——————— (1)$
$Given\ x^2\ +\ y^2\ =\ 400 ——————— (2)$
$From\ (1)\ \hspace 10cm$
$2g_1 = – 10\ \hspace 2cm\ 2f_1 = – 24\ \hspace 2cm\ c_1 = 120$
$g_1 =\ – 5\ \hspace 2cm\ f_1 =\ – 12\ \hspace 2cm\ c_1 = 120$
$Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}$
$C_1 = (5,\ 12)\ \hspace 10cm\ r_1 = \sqrt{(\ -\ 5)^2 + (-\ 12)^2\ -\ 120}$
$\hspace 10cm\ r_1 = \sqrt{25\ +\ 144\ -\ 120}$
$\hspace 10cm\ r_1 = \sqrt{169\ -\ 120}$
$\hspace 10cm\ r_1 = \sqrt{49} =\ 7$
$\boxed{C_1 = ( 5, 12)\ and\ r_1 = 7}$
$From\ (2)\ \hspace 10cm$
$2g_2 = 0\ \hspace 2cm\ 2f_2 = 0\ \hspace 2cm\ c_2 = – 400$
$g_2 =\ 0 \hspace 2cm\ f_2 = 0\ \hspace 2cm\ c_2 = – 400$
$Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}$
$C_2 = (0,\ 0)\ \hspace 10cm\ r_2 = \sqrt{(0)^2 +\ (0)^2 + 400}$
$\hspace 10cm\ r_2 = \sqrt{0\ +\ 0\ +\ 400}$
$\hspace 10cm\ r_2\ =\ \sqrt{400}\ =\ 20$
$\boxed{C_2 = ( 0, 0)\ and\ r_2\ =\ 20}$
$C_1C_2 = \sqrt{(0\ -\ 5)^2 + (0\ -\ 12)^2}$
$C_1C_2 = \sqrt{(-\ 5)^2 + (-\ 12)^2}$
$C_1C_2 = \sqrt{25\ +\ 144}$
$C_1C_2 = \sqrt{169}\ =\ 13$
$r_2 – r_1 \ =\ 20\ -\ 13\ =\ 7\ = C_1C_2$
$\boxed{C_1C_2 = r_2 – r_1}$
$The\ given\ circles\ touch\ each\ other\ internally$