Q.
Vertices of an isosceles triangle of area a2and (−a,0) and (a,0). Equation of the circumcircle of the triangle is
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a
x2+y2+2ax−2ay+a2=0
b
x2+y2−2ax+2ay+a2=0
c
x2+y2=a2
d
none of these
answer is C.
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Detailed Solution
Since the triangle is isosceles, third vertex lies on the line x = 0, perpendicular to the base and passing through the mid-point (0, 0) of the base. As the area is a 2 , distance of the vertex from the base is a as the length of the base is 2a. So vertex of the triangle is (0, ± a) and let the equation of the circle passing through the vertices of the triangle x2+y2+2gx+2fy+c=0, then a2+2ga+c=0and a2±2fa+c=0 ⇒ c=−a2,g=f=0 and the equation of the required circle isx2+y2=a2
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