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

\[\underline{PART\ -\ A}\]

1.  Find the equation of the circle with centre (1, -2) and radius 5 units.

Soln:   We know that  the  equation of circle is (x – h )2 + (y – k )2    =   r2

Here h =  1,  k = – 2  (given)   and   r  =  5

(x – 1 )2  + (y + 2 )2 = 52

x2 – 2x + 1+  y2 +  4y + 4= 25

x2  +   y2   – 2x  + 4y + 5 -25 = 0

x2  +   y2    – 2x  + 4y – 20= 0

Therefore the equation of circle is x2  +   y2    – 2x  + 4y – 20= 0

2.     Find the centre and radius of the circle  x2  +   y2 +  10x  +  8y  +  5 = 0 .

Soln:    Given   x2 +  y2  +  10x  +  8y  +  5 = 0 .

We know the equation of circle is  x2  +   y2  + 2gx  + 2fy  + c = 0

2g  = 10                  2f = 8                  c = 5

g  =  5                     f = 4

centre = ( -g , -f )                              r = √( g2  + f2 – c)

=   ( – 5 , – 4 )                           r = √( (5)2  + (4)2 – 5)

r = √36                r =  6

centre =  ( – 5 , – 4 )      &        r =  6

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

\[4.\ Show\ that\ the\ circles\ x^2\ +\ y^2\ -\ 2\ x\ +\ 4\ y\ -\ 3\ =\ 0\ and\ \hspace{7cm}\]\[x^2\ +\ y^2\ -\ 2\ x\ +\ 4\ y\ +\ 5\ =\ 0\ are\ concentric\ circles\ \hspace{5cm}\]

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

∴The given circles are concentric circles.

U7 Jewelry Special Offer Sale
\[\underline{PART\ -\ B}\]
\[5.\ Show\ that\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ is\ a\ diameter\ of\ the\ circle\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{5cm}\]
\[Soln:\ Given\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{10cm}\]
\[W.\ K.\ T.\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]
\[2\ g\ =\ -\ 6\ \hspace{6cm}\ 2\ f\ =\ 10\ \hspace{3cm}\]
\[g\ =\ -\ 3\ \hspace{3cm}\ f\ =\ 5\ \hspace{3cm}\]
\[centre\ =\ (-\ g,\ -\ f)\ \hspace{8cm}\]
\[ =\ (3,\ -\ 5)\ \hspace{8cm}\]
\[Put\ x\ =\ 3,\ y\ =\ -\ 5\ in\ \hspace{8cm}\]
\[2\ (3)\ +\ 3\ (-5)\ +\ 9\ =\ 0\ \hspace{8cm}\]
\[6\ -\ 15\ +\ 9\ =\ 0\ \hspace{8cm}\]
\[-\ 9\ +\ 9\ =\ 0\ \hspace{8cm}\]
\[0\ =\ 0\ \hspace{8cm}\]
\[\therefore\ (3, -\ 5)\ lies\ on\ the\ line\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ \hspace{5cm}\]
\[\therefore\ 2\ x\ +\ 3\ y\ +\ 9\ =\ 0\ is\ the\ diameter\ of\ the\ circle\ \hspace{5cm}\]

6.     Find the equation of the circle passing through the point (1 , 1 ) and concentric to the circle

x2  +   y2  +  4x  +  6y  –  15 = 0.

Soln:    Equation of concentric circle be  x2  +   y2  +  4x  +  6y  +  k = 0   ………….. ( 1 )

Put  x = 1,  y = 1  in  ( 1 )

(1)2 +  (1)2   + 4 ( 1 )  +  6( 1 ) + k = 0.

1+  1 + 4 + 6+ k = 0.

12 + k = 0.

K  =  – 12

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

\[7.\ Find\ the\ equation\ of\ the\ circle\ concentric\ with\ the\ circle\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 1\ =\ 0\ \hspace{7cm}\]\[and\ passing\ through\ the\ point\ (1,1)\ \hspace{5cm}\]
\[Soln:\ Equation\ of\ concentric\ circle\ be\ x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ +\ k\ =\ 0\ ———- (1)\ \hspace{6cm}\]
\[Put\ x\ = 1,\ y\ =\ 1\ in\ equation\ ( 1 )\ \hspace{10cm}\]
\[(1)^2\ +\ (1)^2\ -\ 6(1)\ +\ 10(1)\ +\ K\ =\ 0\ \hspace{10cm}\]
\[1\ +\ 1\ -\ 6\ +\ 10\ +\ K\ =\ 0\ \hspace{10cm}\]
\[6\ +\ K\ =\ 0\ \hspace{10cm}\]
\[K\ =\ -\ 6\ \hspace{10cm}\]
\[\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]
\[\boxed{x^2\ +\ y^2\ -\ 6\ x\ +\ 10\ y\ -\ 6\ =\ 0}\ \hspace{5cm}\]
\[8.\ Find\ the\ equation\ of\ the\ circle\ concentric\ with\ the\ circle\ x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ -\ 23\ =\ 0\ \hspace{7cm}\]\[and\ having\ radius\ 3\ units\ \hspace{5cm}\]
\[Sol:\ As\ we\ know\ that\ centres\ of concentric\ circles\ is\ same, \hspace{15cm}\]\[we\ will\ find\ the\ centre\ for\ x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ -\ 23\ =\ 0\ \hspace{5cm}\]
\[W.\ K.\ T.\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}\]
\[2\ g\ =\ 8\ \hspace{6cm}\ 2\ f\ =\ -\ 4\ \hspace{3cm}\]
\[g\ =\ 4\ \hspace{3cm}\ f\ =\ -\ 2\ \hspace{3cm}\]
\[centre\ =\ (-\ g,\ -\ f)\ \hspace{8cm}\]
\[ =\ (-\ 4,\ 2)\ \hspace{8cm}\]
\[We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}\]
\[Here\ h\ =\ -\ 4,\ k\ =\ 2\ and\ r\ =\ 3\ (Given)\ \hspace{10cm}\]
\[(x\ +\ 4)^2\ +\ (y\ -\ 2)^2\ =\ 3^2\ \hspace{10cm}\]
\[x^2\ +\ 8\ x\ +\ 16\ +\ y^2\ -\ 4y\ +\ 4\ =\ 9\ \hspace{10cm}\]
\[x^2\ +\ y^2\ +\ 8\ x\ -\ 4y\ +\ 16\ +\ 4\ -\ 9\ =\ 0\ \hspace{10cm}\]
\[x^2\ +\ y^2\ +\ 8\ x\ -\ 4y\ +\ 11\ =\ 0\ \hspace{10cm}\]
\[\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]
\[\boxed{x^2\ +\ y^2\ +\ 8\ x\ -\ 4\ y\ +\ 11\ =\ 0}\ \hspace{5cm}\]
\[\underline{PART\ -\ C}\]
\[9.\ Find\ the\ equation\ of\ the\ circle\ passing\ through\ the\ point\ (-7.1)\ \hspace{7cm}\]\[and\ having\ its\ centre\ at\ (-4,-3)\ \hspace{5cm}\]
\[Soln:\ r\ =\ \sqrt{(-\ 4\ +\ 1)^2\ +\ (-3\ -\ 1)^2}\ \hspace{15cm}\]
\[=\ \sqrt{(3)^2\ +\ (-\ 4)^2}\ \hspace{10cm}\]
\[=\ \sqrt{9\ +\ 16}\ \hspace{10cm}\]
\[r=\ \sqrt{25}\ =\ 5\ \hspace{10cm}\]
\[We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}\]
\[Here\ h\ =\ -\ 4,\ k\ =\ -\ 3\ (Given)\ and\ r\ =\ 5\ \hspace{10cm}\]
\[(x\ +\ 4)^2\ +\ (y\ +\ 3)^2\ =\ 5^2\ \hspace{10cm}\]
\[x^2\ +\ 8\ x\ +\ 16\ +\ y^2\ +\ 6y\ +\ 9\ =\ 25\ \hspace{10cm}\]
\[x^2\ +\ y^2\ +\ 8\ x\ +\ 6y\ +\ 25\ -\ 25\ =\ 0\ \hspace{10cm}\]
\[x^2\ +\ y^2\ +\ 8\ x\ +\ 6y\ =\ 0\ \hspace{10cm}\]
\[\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]
\[\boxed{x^2\ +\ y^2\ +\ 8\ x\ +\ 6\ y\ =\ 0}\ \hspace{5cm}\]
\[10.\ Find\ the\ equation\ of\ the\ circle\ two\ of\ whose\ diameters\ are\ \hspace{15cm}\]\[2x\ -\ 3y\ +\ 1\ =\ 0\ and\ x\ +\ 2y\ -\ 17\ =\ 0\ and\ its\ radius\ is\ 8\ units\ \hspace{10cm}\]
\[Soln:\ 2x\ -\ 3y\ =\ -\ 1\ ————–\ (1)\ \hspace{15cm}\]
\[x\ +\ 2y\ =\ 17\ ————–\ (2)\ \hspace{15cm}\]
\[(2) \times\ 2\ and\ (1)\ -\ (2)\ \implies\ -\ 7y\ =\ -\ 35\ \hspace{10cm}\]
\[\therefore\ \boxed{y\ =\ 5}\ \hspace{10cm}\]
\[Substitute\ y\ =\ 5\ in\ (1)\ \hspace{10cm}\]
\[2\ x\ -\ 15\ =\ -1\ \hspace{10cm}\]
\[2\ x\ =\ 14\ \hspace{10cm}\]
\[\therefore\ \boxed{x\ =\ 7}\ \hspace{10cm}\]
\[Centre\ =\ (7,\ 5)\ \hspace{10cm}\]
\[We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}\]
\[Here\ h\ =\ 7,\ k\ =\ 5\ and\ r\ =\ 8\ \hspace{10cm}\]
\[(x\ -\ 7)^2\ +\ (y\ -\ 5)^2\ =\ 8^2\ \hspace{10cm}\]
\[x^2\ -\ 14\ x\ +\ 49\ +\ y^2\ -\ 10y\ +\ 25\ =\ 64\ \hspace{10cm}\]
\[x^2\ +\ y^2\ -\ 14\ x\ -\ 10y\ +\ 74\ -\ 64\ =\ 0\ \hspace{10cm}\]
\[x^2\ +\ y^2\ -\ 14\ x\ -\ 10y\ +\ 10 =\ 0\ \hspace{10cm}\]
\[\therefore\ The\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]
\[\boxed{x^2\ +\ y^2\ -\ 14\ x\ -\ 10\ y\ +\ 10\ =\ 0}\ \hspace{5cm}\]
10% Off $159+ Code: mom10 at Wayrates.com! Redmagic WW
\[11.\ Show\ that\ the\ circles\ x^2 + y^2 + 2x – 8 = 0\ and\ x^2 + y^2 – 6x + 6y -46 = 0\ touch\ each\ other.\ \hspace10cm\]

Soln:

\[Given\ x^2 + y^2 + 2x – 8 = 0 ——————— (1)\]
\[Given\ x^2 + y^2 – 6x + 6y -46 = 0 ——————— (2)\]
\[From\ (1)\ \hspace 10cm\]
\[2g_1 = 2\ \hspace 2cm\ 2f_1 = 0\ \hspace 2cm\ c_1 = – 8\]
\[g_1 = 1\ \hspace 2cm\ f_1 = 0\ \hspace 2cm\ c_1 = – 8\]
\[Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}\]
\[ C_1 = (- 1,\ 0)\ \hspace 10cm\ r_1 = \sqrt{1^2 + 0^2 + 8}\]
\[ \hspace 10cm\ r_1 = \sqrt{1 + 0 + 8}\]
\[ \hspace 10cm\ r_1 = \sqrt{9} = 3\]
\[\boxed{C_1 = ( -1, 0)\ and\ r_1 = 3}\]
\[From\ (2)\ \hspace 10cm\]
\[2g_2 = -6\ \hspace 2cm\ 2f_2 = 6\ \hspace 2cm\ c_2 = – 46\]
\[g_2 = -3\ \hspace 2cm\ f_2 = 3\ \hspace 2cm\ c_2 = – 46\]
\[Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}\]
\[ C_2 = (3,\ -3)\ \hspace 10cm\ r_2 = \sqrt{(-3)^2 + 3^2 + 46}\]
\[ \hspace 10cm\ r_2 = \sqrt{9 + 9 + 46}\]
\[ \hspace 10cm\ r_2 = \sqrt{64} = 18\]
\[\boxed{C_2 = ( 3, -3)\ and\ r_2 = 8}\]
\[C_1C_2 = \sqrt{(-1-3)^2 + (0- (-3))^2}\]
\[C_1C_2 = \sqrt{(-4)^2 + (3)^2}\]
\[C_1C_2 = \sqrt{16 + 9}\]
\[C_1C_2 = \sqrt{25} = 5\]
\[r_2 – r_1 = 8 – 3 = 5 = C_1C_2\]
\[\boxed{C_1C_2 = r_2 – r_1}\]
\[The\ given\ circles\ touch\ each\ other\ internally\]

12. Prove that the circles  x2  +   y2   – 2x  +  6y + 6= 0 and  x2  +   y2   – 5x  + 6y + 15 = 0 touch  each other.

Soln:    Given x2  +   y2   – 2x  +  6y + 6 = 0 ––––– (1)      and

x2  +   y2 – 5x  + 6y + 15 = 0  ––––– (2)

From ( 1 )

2g1  = -2                    2f1 = 6                c1 = 6

g1=  -1                       f1= 3

centre  is  C1  = ( – g1 , – f1 )                 r1 = √( g12  + f12 – c)

=   ( 1 , -3 )                     r= √( (-1)2  + (3)2 – 6)

r= √( 1  + 9  – 6)

r1 = √4 = 2

C1=  ( 1 , -3 )   &  r1 = 2

From ( 2 )

2 g2  = – 5                    2 f2 = 6                c2 = 15

g2 =  -5/2                      f2 = 3

centre  is  C2  = ( – g2 , – f2 )                 r2 = √( g22  + f22 – c)

=   ( -5/2 , -3 )                r= √( ( -5/2 )2  + (3)2  – 15)

r= √( 25/4  + 9  – 15 )

r= √( 25/4  – 6 )

r= √ (25 – 24)/4

r= √ 1/4 = 1/2

C2=  ( -5/2 , -3 )    &  r2 =   1/2

C1 C2   =  √(( 5/2 – 1 )2 + (-3 + 3 )2)

=   √( 5 – 2)/2 )2 

=   √( 3/2 )2

C1 C2   = 3/2

r1  –  r2    =  2  –   1/2

= (4 – 1)/2

= 3/2 = C1 C2

∴The circles touch each other internally.

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