Find the equation of the circle drawn on the intercept made by the line 3 x + 4 y =12

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Solution

$\begin{matrix} E q u a t i o n o f c i r c l e w i t h e n d p o i n s o f d i a m e t e r . (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0 T h e g i v e n e q u a t i o n o f l i n e 3 x + 4 y = 12 a t y - a x i s, x = 0 ∴ y = 3 \Rightarrow a (x_{1}, y_{1}) = (0, 3) s i m i l a r l y a t x - a x i s y = 0 ∴ x = 4 \Rightarrow B (x_{2}, y_{2}) = (4, 0) u sin g t h e s e i n a b o v e e q u a t i o n o f c i r c l e w e g e t (x - 0) (x - 4) + (y - 3) (y - 0) = 0 ∴ x^{2} - 4 x + y^{2} - 3 y = 0 I t i s t h e r e q u i r e d s o l u t i o n \end{matrix}$

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(- 4, 3)

(12, - 1)

Find the equation of the circle whose diameter is the line segment joining (-4, 3) and (12, -1). Find also the intercept made by it on y-axis.

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