Question

If a circle $C$ passing through the point $(4,0)$ touches the circle $x^2 + y^2 + 4x - 6y = 12$ externally at the point $(1, -1)$, then the radius of $C$ is :

• A. $\sqrt{57}$
• B. $4$
• C. $2 \sqrt{5}$
• D. $5$

Answer 1

Answer: (D)

$x^2 + y^2 + 4x - 6y - 12 = 0$
Equation of tangent at $(1, -1)$
$x - y + 2(x + 1) - 3(y - 1) - 12 = 0$
$3x - 4y - 7 = 0$
$\therefore $ Equation of circle is
$(x^2 + y^2 + 4x - 6y - 12) + \lambda (3x - 4y - 7) = 0$
It passes through $(4, 0)$ :
$(16 + 16 - 12) + \lambda(12 - 7) = 0 $
$\Rightarrow \; 20 + \lambda (5) = 0$
$\Rightarrow \; \lambda = - 4$
$ \therefore \; (x^2 + y^2 + 4x - 6y - 12) - 4(3x - 4y - 7) = 0$
or $x^2 + y^2 - 8x + 10y + 16 = 0$
Radius = $\sqrt{16 + 25 - 16 } = 5 $