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Standard XII

Mathematics

Definition of Circle

The circle ...

Question

The circle $O A B$ where $O$ is origin and $A, B$ are points on the co-ordinate axes is drawn such that the distances of points $A$ and $B$ from the tangent to the circle at origin are $p$ and $q$ respectively. Prove that the diameter of the circle is $p + q$ and its centre is $[\frac{1}{2} \sqrt{p (p + q)}, \frac{1}{2} \sqrt{q (p + q)}]$ .

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Solution

Circle $O A B$ is $x^{2} + y^{2} - a x - b y = 0$
Tangent at $(0, 0)$ is $a x + b y = 0$
$∴ p = \frac{a^{2}}{\sqrt{(a^{2} + b^{2})}}, q = \frac{b^{2}}{\sqrt{(a^{2} + b^{2})}}$
$∴ p + q = \sqrt{(a^{2} + b^{2})} = d i a m e t e r A B$
$∴ a^{2} = p \sqrt{(a^{2} + b^{2})} = p (p + q)$ ,
$b^{2} = q (p + q)$
$∴$ Centre is $(\frac{1}{2} a, \frac{1}{2} b)$ etc.

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The circle OAB where O is origin and A,B are points on the co-ordinate axes is drawn such that the distances of points A and B from the tangent to the circle at origin are p and q respectively. Prove that the diameter of the circle is p+q and its centre is [12√p(p+q),12√q(p+q)].

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