CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
π3
loader
C
π6
loader
D
π2
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image