Question

The equation of the circle having the lines $x ( x - 4 ) + y ( y - 3 ) = 0$ is

• A. $x ^ { 2 } + y ^ { 2 } + 3 x - 6 y - 40 = 0$
• B. $x ^ { 2 } + y ^ { 2 } + 6 x - 3 y - 45 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 8 x + 4 y - 20 = 0$
• D. $x ^ { 2 } + y ^ { 2 } + 4 x + 8 y + 20 = 0$

Answer 1

Answer: (B)

$x ^ { 2 } + y ^ { 2 } + 6 x - 3 y - 45 = 0$

Answer 2

Given pair of normals is $x ^ { 2 } + 2 x y + 3 x + 6 y = 0$ or

$( x + 2 y ) ( x + 3 ) = 0$

$\therefore$ Normals are $x + 3 = 0$

The point of intersection of normals $x + 3 = 0$ is the centre of required circle, we get centre $x ( x - 4 ) + y ( y - 3 ) = 0$ or

$x ^ { 2 } + y ^ { 2 } - 4 x - 3 y = 0$ …..(i)

Its centre $C _ { 2 } = ( 2,3 / 2 )$ and radius $r = \sqrt { 4 + \frac { 9 } { 4 } } = \frac { 5 } { 2 }$

Since the required circle just contains the given circle (i), the given circle should touch the required circle internally from inside. Therefore, radius of the required circle $= \left| C _ { 1 } - C _ { 2 } \right| + r$ $= \sqrt { ( - 3 - 2 ) ^ { 2 } + \left( \frac { 3 } { 2 } - \frac { 3 } { 2 } \right) ^ { 2 } } + \frac { 5 } { 2 }$ $= 5 + \frac { 5 } { 2 } = \frac { 15 } { 2 }$

Hence, equation of required circle is

$( x + 3 ) ^ { 2 } + \left( y - \frac { 3 } { 2 } \right) ^ { 2 } = \left( \frac { 15 } { 2 } \right) ^ { 2 }$ or $x ^ { 2 } + y ^ { 2 } + 6 x - 3 y - 45 = 0$