Question

Tangents are drawn to the circle $x^2+y^2=16$ at the points where it intersects the circle $x^2+y^2-6x-8y-8=0,$ then the point of intersection of these tangents is

• A. $\left(4,\frac{16}{3}\right)$
• B. $\left(12,16\right)$
• C. $\left(3,4\right)$
• D. $\left(16,12\right)$

Answer 1

Answer: (B)

$S_1:x^2+y^2-16=0$
$S_2:x^2+y^2-6x-8y-8=0$
In this case, the common chord is the same as the chord of contact $AB$
So, the equation of $AB$ is $3x+4y-4=0$ which is identical with $x{x}_1+y{y}_1-16=0$
$\frac{x_1}{3}=\frac{y_1}{4}=\frac{16}{4}$
$\Rightarrow x_1=12,y_1=16$
$\Rightarrow P\left(x_1,y_1\right)=\left(12,16\right)$