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  •  The equation of the circle which has diameter as the chord common to circles  <img alt="\mathrm{x^2+y^2-a x=0 \text { and } x^2+y^2-by=0}" src="https://entrancecorner.oncodecogs.com/gif.

 The equation of the circle which has diameter as the chord common to circles  \mathrm{x^2+y^2-a x=0 \text { and } x^2+y^2-by=0} is \mathrm{\left(a^2+b^2\right)\left(x^2+y^2\right)=k(b x+a y)} , where k =

Option: 1

 

a b


Option: 2

\frac{a}{b}


Option: 3

a+b


Option: 4

a^2-b^2


Answers (1)

best_answer

 Let the equation of the circle be 

\mathrm{S}+\mathrm{kS}^{\prime}=0 \quad \text { i.e. } \mathrm{x}^2+\mathrm{y}^2-\mathrm{ax}+\mathrm{k}\left(\mathrm{x}^2+\mathrm{y}^2-\mathrm{by}\right)=0

\text { Centre } \equiv\left(\frac{\mathrm{a}}{2(1+\mathrm{k})}, \frac{\mathrm{bk}}{2(1+\mathrm{k})}\right) \ldots \ldots \text { (A) }

(A) Equation of the common chord of the given circles is S – S′ = 0 i.e. ax – by = 0. 

(A) lies on this chord (diameter)

(B) \begin{aligned} & \quad \therefore \frac{a \cdot a}{2(1+k)}-\frac{b \cdot b k}{2(1+k)}=0 \text {. } \\ & \therefore \mathrm{k}=a^2 / b^2 . \end{aligned}

∴The equation of the circle is  \mathrm{\left(a^2+b^2\right)\left(x^2+y^2\right)-a b(b x+a y)=0}

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Sayak

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