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

This is a Short Answer Type Questions as classified in NCERT ExemplarSol:Sol:Let a be the radius of the circle.Centre of the circle=(−a,−a) Distance of the line 3x−4y+8=0From the centre=Radius of the circle |−3a+4a+8(3)2+(−4)2|=a⇒ |a+85|=32⇒ ±(a+85)=a⇒ a+85=a and −(a+85)=a⇒ a=5a−8⇒ 4a=8 ⇒a=2and a+85=−a ⇒a+8=−5a⇒ 6a=−8 ⇒a=−43∴ a=2 and a≠−43∴ The equation of the circle is (x+2)2+(y+2)2=(2)2⇒x2+4x+4+y2+4y+4=4⇒ x2+y2+4x+4y+4=0Hence, the required equation of the circle is x2+y2+4x+4y+4=0.

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Maths Class 11th Conic Sections Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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

2 months ago

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

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