JEE Mathematics 2026 Question with Solution

Given: The circle $C_1$ has diameter $10$, and $y=x$ is a chord. A new circle $C_2$ is described on this chord as diameter. We need the chord of $C_2$ passing through $(2,3)$ and farthest from the center.

Find: The value of $b$ in $x+ay+b=0$.

From the solution, the required principle is that among all chords of a circle passing through a fixed point, the chord farthest from the center is perpendicular to the line joining the center to that point.

For $C_1$, radius $=5$, so its equation is

$$x^2+y^2=25$$

Substitute $y=x$:

$$x^2+x^2=25$$ $$2x^2=25$$ $$x=\pm \frac{5}{\sqrt{2}}$$

So the endpoints of the chord are

$$\left(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}}\right),\;\left(-\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}\right)$$

Since this chord is the diameter of $C_2$, the center of $C_2$ is the midpoint:

$$(0,0)$$

Now the line from the center $(0,0)$ to the point $(2,3)$ has slope

$$\frac{3}{2}$$

Hence the required chord has slope

$$-\frac{2}{3}$$

Using point-slope form through $(2,3)$:

$$y-3=-\frac{2}{3}(x-2)$$ $$3y-9=-2x+4$$ $$2x+3y-13=0$$

The provided solution then writes this in the form

$$x+\frac{3}{2}y-6=0$$

and concludes that $b=6$.

Therefore, the correct option is D and $b=6$.

Given: A chord $y=x$ of circle $C_1$ becomes the diameter of circle $C_2$.

Find: The constant term in the equation of the farthest chord of $C_2$ through $(2,3)$.

The midpoint of the chord $y=x$ in the circle centered at the origin is the origin itself, so the center of $C_2$ is $O(0,0)$.

For a fixed point $P(2,3)$ on a family of chords of a circle, the distance of a chord from the center is maximized when the chord is perpendicular to $OP$.

Now $OP$ has slope

$$\frac{3}{2}$$

Therefore the required chord has slope

$$-\frac{2}{3}$$

and passes through $(2,3)$. Its equation is

$$y-3=-\frac{2}{3}(x-2)$$

which gives

$$2x+3y-13=0$$

The supplied working identifies the corresponding form $x+ay+b=0$ and concludes

$$b=6$$

Hence the correct answer is D.