A (1,0) and B(7,0) are two points on the axis of x. A point P is taken in the first quadrant such that PAB is an isosceles triangle and PA=PB=5 unit. Find the equation of the circle described on PA as diameter

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Solution

$\begin{matrix} u s i n g d i s t a n c e f o r m u l a \sqrt{{(x - 1)}^{2} + y^{2}} = \sqrt{{(x - 7)}^{2} + y^{2}} (s q u a r i n g b o t h s i d e s) x^{2} + 1 - 2 x = x^{2} + 49 - 14 x N o w, \sqrt{{(4 - 1)}^{2} + y^{2}} = 5 \Rightarrow 9 + y^{2} = 25 ∴ y = \pm 4 S i n c e, p o i n t i s i n f i r s t q u a d r a n t s o y c a n n o t b e n e g a t i v e c o - o r d i a n t e o f p o i n t p = (4, 4) E q u a t i o n o f c i r c l e \Rightarrow (x - 1) (x - 4) + (y - 0) (y - 4) = 0 \Rightarrow x^{2} - 5 x + 4 + y^{2} - 4 y = 0 \Rightarrow x^{2} + y^{2} - 5 x - 4 y + 4 = 0 w h i c h i s t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e . \end{matrix}$

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