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 . If the circle lies below x-axis, then the ordered pair () is equal to [2025]

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 $x^{2} + y^{2} - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair ( $2 a, b^{2}$ ) is equal to [2025]

Smart Achievers · Accepted Answer

(4)

Let, r be the radius of the circle,

$r = \sqrt{\frac{α^{2}}{4} + \frac{β^{2}}{4} - γ} = \frac{β}{2}$

$\Rightarrow \frac{α^{2}}{4} - γ = 0$

$\Rightarrow α^{2} = 4 γ; \frac{α}{2} = a \Rightarrow α = 2 a$

Now, length of intercept on y-axis = $b = 2 \sqrt{\frac{β^{2}}{4} - γ}$

$\Rightarrow \frac{β^{2}}{4} - γ = \frac{b^{2}}{4} \Rightarrow b^{2} = β^{2} - 4 γ$

$∴$ Points ( $2 a, b^{2}$ ) = $(α, β^{2} - 4 γ)$

Solution

Q.
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 $x^{2} + y^{2} - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair ( $2 a, b^{2}$ ) is equal to [2025]

1 $(γ, β^{2} + 4 α)$

2 $(γ, β^{2} - 4 α)$

3 $(α, β^{2} + 4 γ)$

4 $(α, β^{2} - 4 γ)$

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Solution

Q. 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 x2+y2–αx+βy+γ=0. If the circle lies below x-axis, then the ordered pair (2a,b2) is equal to [2025]

1 (γ,β2+4α)

2 (γ,β2–4α)

3 (α,β2+4γ)

4 (α,β2–4γ)