Find the equation of a circle concentric with the circle $x^{2} + y^{2} - 6 x + 12 y + 15 = 0$ and has double its area.

[Hint: Concentric circles have the same center.]

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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:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} - 6 x + 12 y + 15 = 0 \dots (i) \\ C e n t r e = (- g, - f) = (3, - 6) [\begin{array}{l} ? 2 g = - 6 \Rightarrow g = - 3 \\ 2 f = 12 \Rightarrow f = 6 \end{array}] \\ Since t h e c i r c l e i s c o n c e n t r i c w i t h t h e g i v e n c i r c l e \\ ∴ C e n t r e = (3, - 6) \\ N o w l e t t h e r a d i u s o f t h e c i r c l e i s r \\ ∴ r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{9 + 36 - 15} = \sqrt[]{30} \\ A r e a o f t h e g i v e n c i r c l e (i) = π r^{2} = 30 π s q . u n i t \\ A r e a o f t h e r e q u i r e d c i r c l e = 2 * 30 π = 60 π s q . u n i t \\ I f r_{1} b e t h e r a d i u s o f t h e r e q u i r e d c i r c l e \\ π r_{1}^{2} = 60 π \Rightarrow r_{1}^{2} = 60 \\ S o, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y + 6)}^{2} = 60 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 36 + 12 y - 60 = 0 \\ \Rightarrow x^{2} + y^{2} - 6 x + 12 y - 15 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 6 x + 12 y - 15 = 0 . \end{array}$

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Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

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Alok Kumar Singh

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