Exam

JEE Main

Year

2025

Date

28 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Circle

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Circle

Let the equation of the circle, which touches x -axis at the point...

Let the equation of the circle, which touches $x$-axis at the point $(a, 0), a\gt0$ 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 $\left(2 a, b^2\right)$ is equal to

Step-by-step solution

By pythagoras $\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2$ $r=\sqrt{\frac{4 a^2+b^2}{4}}$ Equation of circle is $(x-\alpha)^2+(y-\beta)^2=r^2$ $x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0$ comparision $x^2+y^2-\alpha x+\beta y+r=0$ $$\begin{array}{r} -\alpha=-2 a, \beta=-2 p, r=a^2 \\ \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ \alpha^2+b^2=4 p^2 \\ \alpha^2+b^2=\beta^2 \end{array}$$ So, \left(2 \mathrm{a}, \mathrm{b}^2\right)=\left(\alpha, \beta^2-4 \mathrm{r}\right)