1.2 – N – Analytical Geometry Exercise Problems With Solutions

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 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

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 Part – B

1.    Find the equation of the circle passing through the point  A (2, 3) and having its centre at C ( 4 , 1).

Soln:                r =  √(( 4 – 2 )2 + (1 – 3 )2)

=  √( (2)2  + (-2)2 )

=  √ ( 4 + 4)

r    =  √8

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

Here h = 4,  k = 1  (given)   and   r  =  √8

(x – 4 )2  + (y – 1 )2 =   (√8) 2

 x2   – 8x + 16+  y2  –  2y + 1 = 8

x2  +   y2  –  8x  – 2y + 17– 8= 0

x2  +   y2  –  8x  – 2y  + 9 = 0

2.     Prove that the circles  x2  +   y2  – 8x  + 6y – 23 = 0  and x2  +   y2  – 2x  – 5y + 16 = 0 cut orthogonally .

Soln:    Given  x2  +   y2   – 8x  + 6y – 23 = 0    ––––– (1)      and

x2  +   y2  – 2x  – 5y + 16 = 0

From ( 1 )

2 g1  = – 8                   2 f1 = 6                c1 = -23

g1=  – 4                       f1= 3

From ( 2 )

2 g2  =2                    2 f2 = – 5              c2 = 16

g2=  -1                        f2= – 5/2

The condition for orthogonally is

2g1g2  + 2 f1 f2  =  c1 + c

2(-4)(-1) + 2(3)(-5/2) =  -23 + 16

8   – 15        =      – 7

-7 = -7

∴ The circles cut orthogonally.

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Part – C

1.   Prove that the circles  x2  +   y2  – 4x  +  6y  + 8 = 0 and x2  +   y– 10x  –  6y  +14 = 0 touch each  other. 

Soln:    Given  x2  +   y2   – 4x  +  6y  + 8 = 0 ––––– (1)      and

x2  +   y– 10x  –  6y  +14 = 0  ––––– (2)

From ( 1 )

2g1  = -4                    2f1 = 6               c1 = 8

g1 =  -2                       f1= 3

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

=   ( 2 , -3 )                 r= √( (-2)2  + (3)2 – 8)

r= √( 4  + 9  – 8 )

r1 = √13 – 8 = √5

From ( 2  )

2g2   = -10                    2f2 = -6                c2 = 14

g2 =  – 5                          f2= – 3

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

=   ( 5  , 3 )                  r= √( (-5)2  + (-3)2  – 14)

r= √( 25  + 9  – 14 )

r2 = √20 =   √(4 × 5)  =   2√5

C2=   ( 5   , 3 )       &   r2 =   2√5

  C1C2   =  √(( 5 – 2 )2 + (3 + 3 )2)

=   √( (3)2  + (6)2 )

= √( 9  + 36)

= √45 =   √(9 × 5)   =   3√5

C1C2   =   3√5

r1  +  r2  =   √5   +   2√5

=   3√5   =   C1C2

∴The circles touch each other externally.

Redmagic WW
\[2.\ Show\ that\ the\ circles\ x^2 + y^2 + 2x – 8 = 0\ and\ x^2 + y^2 – 6x + 6y -46 = 0\ touch\ each\ other.\ \hspace10cm\]

Soln:

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

3. 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.

Banggood WW