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Question: The abscissas of two points A and B are the roots of the equations \(x^{2} + 2ax - b^{2} = 0\) and t...

The abscissas of two points A and B are the roots of the equations x2+2ax−b2=0x^{2} + 2ax - b^{2} = 0 and their ordinates are the roots of the circle with AB as diameter is

A

(a2+b2+p2+q2)\sqrt{\left( a^{2} + b^{2} + p^{2} + q^{2} \right)}

B

(a2+p2)\sqrt{\left( a^{2} + p^{2} \right)}

C

(b2+q2)\sqrt{\left( b^{2} + q^{2} \right)}

D

None of these

Answer

(a2+b2+p2+q2)\sqrt{\left( a^{2} + b^{2} + p^{2} + q^{2} \right)}

Explanation

Solution

Let co-ordinates of A and B are (α, β) and ( γ, δ) respectively. α+γ=−2a,αγ=−b2\alpha + \gamma = - 2a,\alpha\gamma = - b^{2} and

β+δ=−ap,βδ=−q2\beta + \delta = - ap,\beta\delta = - q^{2}

Equation of circle with AB as diameter.

(x−α)(x−γ)+(y−β)(y−δ)=0(x - \alpha)(x - \gamma) + (y - \beta)(y - \delta) = 0

⇒x2+y2−x(α+γ)−y(β+δ)+αγ+βδ=0x^{2} + y^{2} - x(\alpha + \gamma) - y(\beta + \delta) + \alpha\gamma + \beta\delta = 0

⇒x2+y2+2ax+2py−b2−q2=09x^{2} + y^{2} + 2ax + 2py - b^{2} - q^{2} = 09

∴\thereforeRadius = (a2+p2+b2+q2)\sqrt{\left( a^{2} + p^{2} + b^{2} + q^{2} \right)}

=(a2+b2+p2+q2)\sqrt{\left( a^{2} + b^{2} + p^{2} + q^{2} \right)}