Question

Tangents are drawn from a point on the circle $x^2+y^2-4x+6y-37=0$ to the circle $x^2+y^2-4x+6y-12=0$. The angle between the tangents is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$x^2+y^2-4x+6y-37=0$
$(x-2{)}^2+(y+3{)}^2-4-9-37=0$
$(x-2{)}^2+(y+3{)}^2=50$

And

$x^2+y^2-4x+6y-12=0$
$(x-2{)}^2+(y+3{)}^2-4-9-12=0$
$(x-2{)}^2+(y+3{)}^2=25$

Hence the above circles are concentric with radius of the larger circle equal to $5\sqrt{2}$ and radius of the smaller one being $5$.

Thus if a point P is chosen on the periphery of the larger circle and two tangents are drawn from the center C to P, then $PC=5\sqrt{2}$.

Let the point of contact be A. Then in the triangle ACP, $\angle A={90}^0$, $AC=5$ and $PC=5\sqrt{2}$.

Thus

$PC\sin \left(\angle APC\right)=AC$

Or

$5\sqrt{2}.\sin \left(\angle APC\right)=5$

$\sin \left(\angle APC\right)=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$

Hence

$\angle APC={45}^0$
Therefore

$\angle P=2\angle APC=2\left({45}^0\right)$
$={90}^0$