Verified Solution • 2025–26 Session moderate

QJEEM25Mab1c8: JEE Main 2025 Mathematics - Class 11 JEEM25Mab1c8 Solution: Let the equation of the circle, which touches -axis at the…

Let the equation of the circle, which touches xx -axis at the point (a, 0), a>0 and cuts off an intercept of length bb on y−axisy-a x i s be x2+y2−αx+βy+γ=0x^2+y^2-\alpha x+\beta y+\gamma=0 . If the circle lies below x−axisx-a x i s , then the ordered pair (2a,b2)\left(2 a, b^2\right) is equal to

Multiple Choice Options

  • A (α,β2+4γ)\left(\alpha, \beta^2+4 \gamma\right)
  • B (α,β2−4γ)\left(\alpha, \beta^2-4 \gamma\right)
  • C (γ,β2−4α)\left(\gamma, \beta^2-4 \alpha\right)
  • D (γ,β2+4α)\left(\gamma, \beta^2+4 \alpha\right)

Step-by-Step Verified Solution

JEE Main 2025 (Online) 28th January Morning Shift Mathematics - Circle Question 1 English Explanation

By pytogorus r2=a2+b24=P2\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2

r=4a2+b24r=\sqrt{\frac{4 a^2+b^2}{4}}

Equation of circle is (x−α)2+(y−β)2=r2(x-\alpha)^2+(y-\beta)^2=r^2

amp;x2+y2−2ax−2py+α2+p2−r2=0amp; comparision x2+y2−αx+βy+r=0amp;−α=−2a,β=−2p,r=a2amp;⇒2a=α,4a2+b2=4p2amp;α2+b2=4p2amp;α2+b2=β2\begin{aligned}& x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0 \\& \text { comparision } x^2+y^2-\alpha x+\beta y+r=0 \\& -\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{aligned}

So, (2a,b2)=(α,β2−4r)\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)

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