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From a point P, two mutually perpendicular tangents are drawn to circles $\mathrm{x^2+y^2=a^2}$ and $\mathrm{x^2+y^2=b^2}$ . Then the locus of P is a circle of radius

Option: 1
$\mathrm{a+b}$

Option: 2
$\mathrm{\sqrt{a^2+b^2}}$

Option: 3
$\mathrm{a^2+b^2}$

Option: 4
$\mathrm{2 \sqrt{a^2+b^2}}$

Answers (1)

best_answer

Equation of a tangent to $\mathrm{x^2+y^2=a^2}$ is -

$\mathrm{x \cos \alpha+y \sin \alpha=a}$ -------------(1)

Equation of a tangent to $\mathrm{ x^2+y^2=b^2}$

$\mathrm{x \cos \beta+y \sin \beta=b}$ --------(2)

As (1) and (2) are perpendicular to each other

$\mathrm{ \left(-\frac{\cos \alpha}{\sin \alpha}\right)\left(-\frac{\cos \beta}{\sin \beta}\right)=-1}$

Or $\mathrm{ \cos (\alpha-\beta)=0 \Rightarrow \alpha-\beta=\pi / 2}$

$\therefore$ The equation of tangent (2) becomes

$\mathrm{ x \cos \left(\alpha-\frac{\pi}{2}\right)+y \sin \left(\alpha-\frac{\pi}{2}\right)=b }$

$\mathrm{ x \sin \alpha-y \cos \alpha=b}$ ----------------(3)

Let P be (x1, y1) which lies on (2) and (3).

$\mathrm{\Rightarrow x_1 \cos \alpha+y_1 \sin \alpha=a x_1 \sin \alpha-y_1 \cos \alpha=b}$

On eliminating $\mathrm{\alpha }$ , we obtain $\mathrm{x_1^2+y_1^2=a^2+b^2 }$

The locus of $\mathrm{\left(x_1, y_1\right) \text { is } x^2+y^2=a^2+b^2 }$

$\therefore$ The radius of the circle is $\mathrm{\sqrt{a^2+b^2} }$

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