Question

The equation of the circle passing through the point $( - 1 , - 3 )$ and touching the line $4 x + 3 y - 12 = 0$ at the point (3, 0), is.

• A. $x ^ { 2 } + y ^ { 2 } - 2 x + 3 y - 3 = 0$
• B. $x ^ { 2 } + y ^ { 2 } + 2 x - 3 y - 5 = 0$
• C. $2 x ^ { 2 } + 2 y ^ { 2 } - 2 x + 5 y - 8 = 0$
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

Let the equation be

$x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y + c = 0$ ….(i)

But it passes through $( - 1 , - 3 )$ and (3, 0) therefore

$10 - 2 g - 6 f + c = 0$ ….(ii)

$9 + 6 g + c = 0$ ….(iii)

Also centre is $C ( - g , - f )$.

Slope of tangents $= - \frac { 4 } { 3 } \Rightarrow$ Slope of normal $= \frac { 3 } { 4 }$

$\Rightarrow \frac { f } { 3 + g } = \frac { 3 } { 4 } \Rightarrow 3 g - 4 f + 9 = 0$ .....(iv)

Now on solving (ii), (iii) and (iv), we get

$g = - 1 , f = \frac { 3 } { 2 }$ and $c = - 3$

Therefore, the equation of circle is

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

Trick : The points (–1, –3) and (3, 0) must satisfy the equation of circle. Circle given in (1) satisfies both the points. Also check whether it touches the line $4 x + 3 y - 12 = 0$ or not.