Find the equation of a circle which touches both axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.
[Hint: Let $a$ be the radius of the circle, then $(- a, - a)$ will be the center, and the perpendicular distance from the center to the given line gives the radius of the circle.]

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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:

Sol:

$\begin{array}{l} L e t a b e t h e r a d i u s o f t h e c i r c l e . \\ C e n t r e o f t h e c i r c l e = (- a, - a) \\ Distance o f t h e l i n e 3 x - 4 y + 8 = 0 \\ F r o m t h e c e n t r e = R a d i u s o f t h e c i r c l e \\ | \frac{- 3 a + 4 a + 8}{\sqrt[]{{(3)}^{2} + {(- 4)}^{2}}} | = a \\ \Rightarrow | \frac{a + 8}{5} | = \frac{3}{2} \\ \Rightarrow \pm (\frac{a + 8}{5}) = a \\ \Rightarrow \frac{a + 8}{5} = a a n d - (\frac{a + 8}{5}) = a \\ \Rightarrow a = 5 a - 8 \\ \Rightarrow 4 a = 8 \Rightarrow a = 2 \\ a n d \frac{a + 8}{5} = - a \Rightarrow a + 8 = - 5 a \\ \Rightarrow 6 a = - 8 \Rightarrow a = - \frac{4}{3} \\ ∴ a = 2 a n d a \neq - \frac{4}{3} \\ ∴ T h e e q u a t i o n o f t h e c i r c l e i s \\ {(x + 2)}^{2} + {(y + 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} + 4 x + 4 + y^{2} + 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} + 4 x + 4 y + 4 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s x^{2} + y^{2} + 4 x + 4 y + 4 = 0 . \end{array}$

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