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# The equation of circumcircle of an equilateral triangle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and one vertex of the triangle is (1,1). The equation of incircle of the triangle is A) $4({x^2} + {y^2}) = {g^2} + {f^2}$B) $4({x^2} + {y^2}) + 8gx + 8fy = (1 - g)(1 + 3g) + (1 - f)(1 + 3f)$C) $4({x^2} + {y^2}) + 8gx + 8fy = {g^2} + {f^2}$D) $4({x^2} + {y^2}) + 8gx + 8fy = {g^2} + fy = (2 - g)(1 + 3g)(2 - f)(2 + 3f)$

Last updated date: 20th Jun 2024
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Hint: The centers of the circumcircle and the incircle coincide in the equilateral triangle. Also if a curve is passing through a point (h,k) the point (h,k) must satisfy the equation of the curve at that point.
Also, remember that the radius of the incircle is half of the radius of the circumcircle.

Formula used:
The general equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ then
Its centre is at $( - g, - f)$, and its radius $R = \sqrt {{{( - g - h)}^2} + {{( - f - k)}^2}}$

In an equilateral triangle, we know that the radius of the incircle is half of the radius of the circumcircle.
$r = \dfrac{R}{2}$ also Its centre is at $( - g, - f)$ .
$\Rightarrow {(x + g)^2} + {(y + f)^2} = {r^2} = \dfrac{{{R^2}}}{4}$
Now putting the value of ${R^2}$
$\Rightarrow {(x + g)^2} + {(y + f)^2}$= $\dfrac{1}{4}{\left( {\sqrt {{{( - g - 1)}^2} + {{( - f - 1)}^2}} } \right)^2}$
Further solving the equation we get,
$\Rightarrow {x^2} + {g^2} + 2xg + {y^2} + {f^2} + 2yf = \dfrac{1}{4}\left( {{g^2} + 1 + 2g + {f^2} + 1 + 2f} \right)$
$\Rightarrow 4({x^2} + {y^2}) + 8xg + 8yf + 4{y^2} + 4{f^2} = \left( {{g^2} + 1 + 2g + {f^2} + 1 + 2f} \right)$
Now try to set the equation in required format,
$\Rightarrow 4({x^2} + {y^2}) + 8xg + 8yf = (1 + 2g - 3{g^2}) + (1 + 2f - 3{f^2})$
On factorisation, we get
$\Rightarrow 4({x^2} + {y^2}) + 8gx + 8fy = (1 - g)(1 + 3g) + (1 - f)(1 + 3f)$

Thus option(B) is correct.

Note:
General equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$,where $g,f,c$ are constant , also the centre of the circle is at $( - g, - f)$ i.e (-coefficient of $\dfrac{x}{2}$, -coefficient of $\dfrac{y}{2}$)
And radius $R = \sqrt {{{( - g - h)}^2} + {{( - f - k)}^2}}$
Also remember the equation of circle in central form as ${(x - h)^2} + {(y - k)^2} = {r^2}$ here ($h,k$) is the centre and $r$ is the radius of the circle.