Circle | JEE Main 2025 | Circle Touching x-Axis & Intercept

JEE MAIN 2025

28-01-2025 SHIFT-1

Question

Let the equation of the circle, which touches x-axis at the point $(a,0),a > 0$ and cuts off an intercept of length b on y-axis be ${x^2} + {y^2} - \alpha x + \beta y + \gamma = 0$. If the circle lies below x-axis, then the ordered pair $\left({2a,{b^2}} \right)$ is equal to

Select the correct option:

$\left( {\alpha ,{\beta ^2} + 4\gamma } \right)$

$\left( {\gamma ,{\beta ^2} - 4\alpha } \right)$

$\left( {\alpha ,{\beta ^2} - 4\gamma } \right)$

$\left( {\gamma ,{\beta ^2} + 4\alpha } \right)$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution