  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
π4  B
π3  C
π6  D
π2  Solution

The correct option is D $$\displaystyle\frac{\pi}{2}$$$$x^{2}+y^{2}-4x+6y-37=0$$$$(x-2)^{2}+(y+3)^{2}-4-9-37=0$$$$(x-2)^{2}+(y+3)^{2}=50$$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 $$PC \sin(\angle APC)=AC$$Or $$5\sqrt{2}.\sin(\angle APC)=5$$$$\sin(\angle APC)=\dfrac{5}{5\sqrt{2}}=\dfrac{1}{\sqrt{2}}$$Hence $$\angle APC=45^{0}$$Therefore $$\angle P=2\angle APC=2(45^{0})$$$$=90^{0}$$Maths

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