Find the equation of the circle having $(1, - 2)$ as its center and passing through the line $3 x + y = 14$ , $2 x + 5 y = 18$ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 3 x + y = 14 \dots (i) \\ a n d 2 x + 5 y = 18 \dots (i i) \\ F r o m e q . (i) w e g e t \\ y = 14 - 3 x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q . (i i) w e g e t \\ \Rightarrow 2 x + 5 (14 - 3 x) = 18 \\ \Rightarrow 2 x + 70 - 15 x = 18 \\ \Rightarrow - 13 x = - 70 + 18 \\ \Rightarrow - 13 x = - 52 \Rightarrow x = 4 \\ F r o m e q . (i i i) w e g e t, y = 14 - 3 * 4 = 2 \\ ∴ point o fintersection i s (4, 2) \\ N o w, r a d i u s r = \sqrt[]{{(4 - 1)}^{2} + {(2 + 2)}^{2}} = \sqrt[]{{(3)}^{2} + {(4)}^{2}} = \sqrt[]{9 + 16} = 5 \\ S o, t h e e q u a t i o n o f c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y + 2)}^{2} = {(5)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} + 4 y + 4 = 25 \\ \Rightarrow x^{2} + y^{2} - 2 x + 4 y - 20 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x + 4 y - 20 = 0 . \end{array}$

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