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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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A circle of radius $\sqrt 20$ has the point $(-1,3)$. Its centre is on the line $x=-y$. Which of the following equations satisfies these conditions?

$(1) x^2+y^2-2x+2y -18=0$

$(2) x^2+y^2+2x-2y+18=0$

$(3) x^2+y^2+10x-10y+30=0$

$(4) x^2+y^2-10x+10y-30=0$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • The general equation of the circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where the centre is $(-g, -f)$
  • The radius of the circle is $\sqrt{g^2+f^2-c}$
Given Radius $ = \sqrt {20} \rightarrow \sqrt{g^2+f^2-c} = \sqrt {20} \rightarrow g^2+f^2-c = 20....(1)$
Given that the circle contains the point $(1,-3) \rightarrow 1^2+(-3)^2 + 2g\times 1 + 2f \times (-3) + c = 0 \rightarrow 2g-6f-c = 10.... (2)$
Given that the centre lies on $x=-y \rightarrow f = -g....(3)$
We have to solve the simultaneous equations $(1),\;(2)$ and $(3)$:
Substituting $(3)$ in $(2)$, we get $2g - 6f - c = 10 \rightarrow 2g - 6(-g) - c = 10 \rightarrow 2g+6g-c = 10 \rightarrow c = 8g-10...(4)$
Substituting $(3)$ and $(4)$ in $(1)$, we get $g^2+(-g)^2 - 8g + 10 = 20$
$\qquad g^2+g^2-8g=10$
$\qquad g^2 - 4g - 5 = 0$
$\qquad (g+1)(g-5) = 0 \rightarrow g = -1$ or $g=5$
Case 1: Assuming $g=-1$
$\Rightarrow f = -g = -1$ and $c = 8g - 10 = 8\times(-1) - 10 = -18$
$\Rightarrow x^2+y^2 + 2(-1)x + 2(1)y - 18 = 0$
$\Rightarrow x^2+y^2-2x+2y-18=0$
Case 2: Assuming $g=5$
$\Rightarrow f = -g = -5$ and $c = 8g-10 = 8 \times (5) -10 = 30$
$\Rightarrow x^2 + y^2 + 2(5)x+2(-f)y+30 = 0$
$\Rightarrow x^2+y^2+10x-10y+30=0$
So, there are two equations that satisfy the given conditions as shown above.
answered Apr 9, 2014 by balaji.thirumalai
 

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