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Mathematics · Circles

JEE Main 2025 — 28 January, Morning Shift — Question 8

Let the equation of the circle, which touches xx-axis at the point (a,0),a>0(a, 0), a>0 and cuts off an intercept of length bb on yy-axis be x2+y2−αx+βy+γ=0x^{2}+y^{2}-\alpha x+\beta y+\gamma=0. If the circle lies below xx-axis, then the ordered pair ( 2a,b22 \mathrm{a}, \mathrm{b}^{2} ) is equal to

  1. Option A:

    (α,β2+4γ)\left(\alpha, \beta^{2}+4 \gamma\right)

  2. Option B:

    (γ,β2−4α)\left(\gamma, \beta^{2}-4 \alpha\right)

  3. Option C:

    (γ,β2+4α)\left(\gamma, \beta^{2}+4 \alpha\right)

  4. Option D:

    (α,β2−4γ)\left(\alpha, \beta^{2}-4 \gamma\right)

    Correct

Answer: D

Step-by-step solution

Let   the   equation   of   the   circle   be   x2+y2−αx+βy+γ=0.Comparing   with   x2+y2+2gx+2fy+c=0, we   have:g=−α2,f=β2,c=γ.∴  Centre   C(α2,−β2),r=α2+β24−γ.The   circle   touches   the   x-axis at (a,0), so   the   point   (a,0) satisfies   the   equation:a2−αa+γ=0⇒γ=αa−a2.Since   the   circle   lies   below   the   x-axis,   its   centre   is   below   the   x-axis,hence   the   radius   r=−β2 (as β>0).Using   r2=α2+β24−γ, we   get:β24=α2+β24−γ⇒α2=4γ⇒γ=α24.Substituting   γ=α24 into   γ=αa−a2, we have:αa−a2=α24⇒4αa−4a2=α2⇒(2a−α)2=0.∴  2a=α.Now,   for   the   y-axis   intercept:   when   x=0,y2+βy+γ=0⇒b=β2−4γ.∴  b2=β2−4γ.Hence,   the   ordered   pair   is:(2a,b2)=(α,β2−4γ).Correct   Option:   (d)\begin{aligned} &\text{Let\; the\; equation\; of\; the\; circle\; be\; } x^{2}+y^{2}-\alpha x+\beta y+\gamma=0. \\[6pt] &\text{Comparing\; with\; } x^{2}+y^{2}+2gx+2fy+c=0, \text{ we\; have:} \\[4pt] &g=-\frac{\alpha}{2}, \quad f=\frac{\beta}{2}, \quad c=\gamma. \\[6pt] &\therefore\; \text{Centre\; } C\left(\frac{\alpha}{2},-\frac{\beta}{2}\right), \quad r=\sqrt{\frac{\alpha^{2}+\beta^{2}}{4}-\gamma}. \\[8pt] &\text{The\; circle\; touches\; the\; } x\text{-axis at } (a,0), \text{ so\; the\; point\; } (a,0) \text{ satisfies\; the\; equation:} \\[4pt] &a^{2}-\alpha a+\gamma=0 \quad \Rightarrow \quad \gamma=\alpha a-a^{2}. \\[8pt] &\text{Since\; the\; circle\; lies\; below\; the\; } x\text{-axis,\; its\; centre\; is\; below\; the\; } x\text{-axis,} \\[4pt] &\text{hence\; the\; radius\; } r=-\frac{\beta}{2} \text{ (as } \beta>0). \\[6pt] &\text{Using\; } r^{2}=\frac{\alpha^{2}+\beta^{2}}{4}-\gamma, \text{ we\; get:} \\[4pt] &\frac{\beta^{2}}{4}=\frac{\alpha^{2}+\beta^{2}}{4}-\gamma \quad \Rightarrow \quad \alpha^{2}=4\gamma \quad \Rightarrow \quad \gamma=\frac{\alpha^{2}}{4}. \\[8pt] &\text{Substituting\; } \gamma=\frac{\alpha^{2}}{4} \text{ into\; } \gamma=\alpha a-a^{2}, \text{ we have:} \\[4pt] &\alpha a-a^{2}=\frac{\alpha^{2}}{4} \quad \Rightarrow \quad 4\alpha a-4a^{2}=\alpha^{2} \quad \Rightarrow \quad (2a-\alpha)^{2}=0. \\[6pt] &\therefore \; 2a=\alpha. \\[10pt] &\text{Now,\; for\; the\; } y\text{-axis\; intercept:\; when\; } x=0, \\[4pt] &y^{2}+\beta y+\gamma=0 \quad \Rightarrow \quad b=\sqrt{\beta^{2}-4\gamma}. \\[6pt] &\therefore \; b^{2}=\beta^{2}-4\gamma. \\[10pt] &\text{Hence,\; the\; ordered\; pair\; is:} \\[4pt] &(2a,b^{2})=(\alpha,\beta^{2}-4\gamma). \\[6pt] &\boxed{\text{Correct\; Option:\; (d)}} \end{aligned}

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Circles
Topic
Introduction to Circles