JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.
Problem
Let the equation of the circle, which touches the x-axis at the point $(a,0)$ ($a>0$) and cuts off an intercept of length $b$ on the y-axis, be $x^2+y^2-\alpha x+\beta y+\gamma=0$. If the circle lies below the x-axis, then the ordered pair $(2a, b^2)$ is equal to:
Key insight. A circle tangent to the x-axis has its centre directly above or below the point of tangency — so the centre’s x-coordinate is exactly $a$, and comparing this to the general-form centre $\left(\frac{\alpha}{2},-\frac{\beta}{2}\right)$ immediately gives $2a=\alpha$. The y-axis intercept length, meanwhile, follows from the standard “chord on an axis” formula applied directly to the circle’s own equation — no need to separately work out the radius or the point of tangency’s exact coordinates.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Compare the general circle equation’s centre, $\left(\frac{\alpha}{2},-\frac{\beta}{2}\right)$, to the tangency condition to relate $\alpha$ and $a$. Then find where the circle crosses the y-axis by substituting $x=0$ into its equation, and use the standard formula for the length of a chord cut on an axis to express $b^2$ directly in terms of $\beta$ and $\gamma$.
Solution
Step 1 — Identify the circle’s centre
For $x^2+y^2-\alpha x+\beta y+\gamma=0$, the centre is at $\left(\dfrac{\alpha}{2},-\dfrac{\beta}{2}\right)$.
Step 2 — Use the tangency condition
Since the circle touches the x-axis at $(a,0)$, the centre must lie directly above or below this point — so the centre’s x-coordinate equals $a$:
$$\frac{\alpha}{2} = a \implies \alpha = 2a$$
Step 3 — Find where the circle meets the y-axis
Substituting $x=0$ into the circle’s equation:
$$y^2+\beta y+\gamma = 0$$
Let the two roots (the y-coordinates where the circle crosses the y-axis) be $y_1, y_2$. By Vieta’s formulas: $y_1+y_2=-\beta$ and $y_1y_2=\gamma$.
Step 4 — Compute the length of the y-axis intercept
The length of the intercept is $b = |y_1-y_2|$:
$$b^2 = (y_1-y_2)^2 = (y_1+y_2)^2-4y_1y_2 = \beta^2-4\gamma$$
Step 5 — Combine both results
$$(2a,\ b^2) = (\alpha,\ \beta^2-4\gamma)$$
Answer
$$(2a,\ b^2) = (\alpha,\ \beta^2-4\gamma)$$
Common mistakes
- Trying to find the exact centre and radius numerically before relating $\alpha,\beta,\gamma$ to $a,b$. Since the problem only asks for $(2a,b^2)$ in terms of the circle’s own coefficients, working directly with the tangency condition and the y-axis intercept formula is far faster than solving for specific numeric coordinates.
- Getting the y-axis intercept-length formula wrong. For a circle’s general equation, the y-axis intercept length is $\sqrt{\beta^2-4\gamma}$ — mixing this up with the x-axis intercept formula $\sqrt{\alpha^2-4\gamma}$ (which uses $\alpha$ instead of $\beta$) leads to the wrong final pair.
Practise next
- Let a circle $x^2+y^2-\alpha x+\beta y+\gamma=0$ touch the y-axis at $(0,c)$ and cut an intercept of length $d$ on the x-axis. Express $(2c, d^2)$ in terms of $\alpha,\beta,\gamma$, using the same approach.
Show answer
$(2c,\ d^2)=\left(-\beta,\ \alpha^2-4\gamma\right)$.
Setting $x=0$ gives $y^2+\beta y+\gamma=0$. Touching the $y$-axis at $(0,c)$ means this has a repeated root $c$, so $2c=-\beta$ and $\gamma=c^2$.
Setting $y=0$ gives $x^2-\alpha x+\gamma=0$, whose two roots are the ends of the $x$-intercept. Their difference squared is $(x_1+x_2)^2-4x_1x_2=\alpha^2-4\gamma$, so $d^2=\alpha^2-4\gamma$.
Tangency becomes ‘repeated root’ and an intercept length becomes ‘difference of roots’ — both read straight off the coefficients.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.