JEE Mathematics 2023 Question with Solution

Given: The triangle is formed by the tangents at $(2, 2)$ to the curves $y^2 = 2x$ and $x^2 + y^2 = 4x$, and the line $x + y + 2 = 0$.

Find: The value of $r^2$, where $r$ is the circumradius of the triangle.

For $y^2 = 2x$, differentiating implicitly:

$$2y\frac{dy}{dx} = 2$$

So,

$$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$$

At $(2, 2)$, the tangent is

$$y - 2 = \frac{1}{2}(x - 2)$$

which gives

$$x - 2y + 2 = 0$$

For $x^2 + y^2 = 4x$, differentiating implicitly:

$$2x + 2y\frac{dy}{dx} = 4$$

So,

$$\frac{dy}{dx} = \frac{2 - x}{y}$$

At $(2, 2)$, the slope is

$$0$$

Hence the tangent is

$$y = 2$$

The third line is

$$x + y + 2 = 0$$

Now find the vertices of the triangle.

Intersection of $x - 2y + 2 = 0$ and $y = 2$:

$$x - 4 + 2 = 0$$

so

$$x = 2$$

Thus one vertex is $(2, 2)$.

Intersection of $y = 2$ and $x + y + 2 = 0$:

$$x + 2 + 2 = 0$$

so

$$x = -4$$

Thus second vertex is $(-4, 2)$.

Intersection of $x - 2y + 2 = 0$ and $x + y + 2 = 0$: Subtracting,

$$-3y = 0$$

so

$$y = 0$$

and then

$$x = -2$$

Thus third vertex is $(-2, 0)$.

Now the side lengths are:

$$AB = 6, \quad AC = \sqrt{(2 + 2)^2 + (2 - 0)^2} = \sqrt{20} = 2\sqrt{5}, \quad BC = \sqrt{(-4 + 2)^2 + (2 - 0)^2} = \sqrt{8} = 2\sqrt{2}$$

Area of the triangle using base $AB = 6$ and height $2$:

$$\Delta = \frac{1}{2} \times 6 \times 2 = 6$$

Using the circumradius formula

$$r = \frac{abc}{4\Delta}$$

we get

$$r = \frac{6 \cdot 2\sqrt{5} \cdot 2\sqrt{2}}{4 \cdot 6} = \sqrt{10}$$

Therefore,

$$r^2 = 10$$

So the required value is $10$.

The solution contains a slope inconsistency, but its final conclusion $r^2 = 10$ is correct.