Let in an equilateral Delta ABC, A(-1 + a cos | vedclass

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(B) In an equilateral triangle,the circumcenter coincides with the centroid. The coordinates of the vertices are given as $A(-1 + a \cos 	heta, 2 + a

Vedclass · Accepted Answer

$9x^2 + 9y^2 + 18x - 36y + 45 - 16b^2 = 0$

Vedclass · Answer

(B) In an equilateral triangle,the circumcenter coincides with the centroid. The coordinates of the vertices are given as $A(-1 + a \cos 	heta, 2 + a \sin 	heta),$ $B(-1 + a \cos \alpha, 2 - a \sin \alpha),$ and $C(-1 + a \sin \beta, 2 + a \cos \beta).$ By observing the structure,the circumcenter is $O(-1, 2).$ The distance from the circumcenter to any vertex is the circumradius $R.$ In an equilateral triangle,the length of the median $m = \frac{3}{2}R.$ Given $m = 2b,$ we have $2b = \frac{3}{2}R \implies R = \frac{4b}{3}.$ The equation of the circle with center $(-1, 2)$ and radius $R = \frac{4b}{3}$ is $(x + 1)^2 + (y - 2)^2 = (\frac{4b}{3})^2.$ Expanding this: $x^2 + 2x + 1 + y^2 - 4y + 4 = \frac{16b^2}{9}.$ Multiplying by $9$: $9x^2 + 9y^2 + 18x - 36y + 45 = 16b^2.$ Wait,re-evaluating the median length: The distance from $A$ to the midpoint of $BC$ is $2b.$ For an equilateral triangle,$R = \frac{2}{3} 	imes 	ext{median} = \frac{2}{3}(2b) = \frac{4b}{3}.$ Actually,the standard relation is $R = \frac{2}{3} 	imes 	ext{median}.$ Thus $R = \frac{4b}{3}.$ $(x+1)^2 + (y-2)^2 = \frac{16b^2}{9} \implies 9x^2 + 9y^2 + 18x - 36y + 45 - 16b^2 = 0.$ Comparing with options,option $B$ is correct.

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