JEE Mathematics 2023 Question with Solution

Given: Common tangents are drawn from a point A on the x-axis to the circle $x^2 + y^2 = 8$ and the parabola $y^2 = 16x$.

Find: The value of $(QR)^2$, where Q and R are the points of contact on the two curves.

Take the common tangent in slope form as

For the circle $x^2 + y^2 = 8$, the perpendicular distance from the centre to the tangent equals the radius. Hence,

Therefore,

So the common tangents are

Point of contact on parabola: let $m = 1$. Using the parametric point of contact form shown in the solution,

we get

Point of contact on circle is

Now,

Therefore, the value of $(QR)^2$ is $72$. The correct option is D. The source solution labels it as option C, but the computed value $72$ matches option D in the given options.

Given: A common tangent touches the circle $x^2 + y^2 = 8$ at Q and the parabola $y^2 = 16x$ at R.

Find: $(QR)^2$.

Write the tangent in slope form for the parabola $y^2 = 4ax$ with $4a = 16$, so $a = 4$:

The extracted solution text displays the tangent in an equivalent slope setup and then uses the tangency condition with the circle to obtain $m = \pm 1$.

Using the circle tangency condition, the solution arrives at

Choosing $m = 1$ gives the relevant contact points reported in the solution:

Hence the squared distance is

So the required value is $72$.