JEE Main 2018 — Circle Question with Solution

NOTE :

Equation of tangent at (x₁, y₁) to the curve x² = 4ay is

xx₁ = 4a $$\left(\frac{y+y_1}{2}\right)$$

Now equation of tangent at (1, 7) to x² = y $$-$$ 6 is

$$x.1=4.\frac{1}{4}\left(\frac{y+7}{2}\right)-6$$

$$\Rightarrow 2x=y+7-12$$

$$\Rightarrow 2x-y+5=0$$

This tangent touches the circle.



So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

For the circle,

$$x^2+y^2+16x+12y+C=0$$

center is ($$-$$8, $$-$$6)

and radius (r) = $$\sqrt{8^2+6^2-c}$$

$$=\sqrt{100-c}$$

Distance of the tangent from the center of the circle

d = $$\left|\frac{2\left(-8\right)-\left(-6\right)+5}{\sqrt{2^2+1^2}}\right|$$

$$=\left|\frac{-16+6+5}{\sqrt{5}}\right|$$

And we know d = r

$$\therefore$$ $$\left|\frac{-16+11}{\sqrt{5}}\right|=\sqrt{100-c}$$

$$\Rightarrow \left|\frac{-5}{\sqrt{5}}\right|=\sqrt{100-c}$$

$$\Rightarrow \left|-\sqrt{5}\right|=\sqrt{100-c}$$

$$\Rightarrow 5=100-c$$

$$\Rightarrow c=95$$