Find the equation of the circle which passes through the points (2, 3) and (4, 5), and its center lies on the straight line $y - 4 x + 3 = 0$ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \dots (i) \\ I f t h e c i r c l e passes t h r o u g h (2, 3) a n d (4, 5) t h e n \\ {(2 - h)}^{2} + {(3 - k)}^{2} = r^{2} \dots (i i) \\ a n d {(4 - h)}^{2} + {(5 - k)}^{2} = r^{2} \dots (i i i) \\ S u b t r a c t i n g e q n . (i i i) f r o m e q n . (i i) w e h a v e \\ {(2 - h)}^{2} - {(4 - h)}^{2} + {(3 - k)}^{2} - {(5 - k)}^{2} = 0 \\ \Rightarrow 4 + h^{2} - 4 h - 16 - h^{2} + 8 h + 9 + k^{2} - 6 k - 25 - k^{2} + 10 k = 0 \\ \Rightarrow 4 h + 4 k - 28 = 0 \\ \Rightarrow h + k = 7 \dots (i v) \\ , t h e c e n t r e (h, k) l i e s o n t h e l i n e y - 4 x + 3 = 0 \\ t h e n k - 4 h - 3 = 0 \Rightarrow k = 4 h - 3 \\ P u t t i n g t h e v a l u e o f k i n e q n . (i v) w e g e t \\ h + 4 h - 3 = 7 \Rightarrow 5 h = 10 \Rightarrow h = 2 \\ F r o m (i v) w e g e t k = 5 \\ P u t t i n g t h e v a l u e o f h a n d k i n e q n . (i i) w e g e t \\ {(2 - 2)}^{2} + {(3 - 5)}^{2} = r^{2} \Rightarrow r^{2} = 4 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 2)}^{2} + {(y - 5)}^{2} = 4 \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 - 10 y = 4 \\ \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 25 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 4 x - 10 y + 25 = 0 . \end{array}$

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