First slide

Definition of a circle

Question

Vertices of an isosceles triangle of area $a^{2}$
and $(- a, 0)$ and $(a, 0) .$ Equation of the circumcircle of the triangle is

Moderate

A

$x^{2} + y^{2} + 2 a x - 2 a y + a^{2} = 0$
B

$x^{2} + y^{2} - 2 a x + 2 a y + a^{2} = 0$
C

$x^{2} + y^{2} = a^{2}$
D

none of these

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 $x^{2} + y^{2} + 2 g x + 2 f y + c = 0,$ then $a^{2} + 2 g a + c = 0$
and $a^{2} \pm 2 f a + c = 0$
$\Rightarrow c = - a^{2}, g = f = 0$ and the equation of the required circle is
$x^{2} + y^{2} = a^{2}$

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