B and C are fixed points having coordinates ( | vedclass

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Question

(A) Let the coordinates of vertex $A$ be $(a, b)$ and the centroid $G$ be $(h, k)$.Since $\angle BAC = 90^o$ and $B(3, 0), C(-3, 0)$,the point $A$ lie

Vedclass · Accepted Answer

$x^2 + y^2 = 1$

Vedclass · Answer

(A) Let the coordinates of vertex $A$ be $(a, b)$ and the centroid $G$ be $(h, k)$.Since $\angle BAC = 90^o$ and $B(3, 0), C(-3, 0)$,the point $A$ lies on a circle with diameter $BC$. The center of this circle is the origin $(0, 0)$ and the radius is $3$. Thus,the equation of the circle is $x^2 + y^2 = 3^2 = 9$. Therefore,$a^2 + b^2 = 9$.The centroid $G(h, k)$ of $\Delta ABC$ is given by $h = \frac{a + 3 - 3}{3} = \frac{a}{3}$ and $k = \frac{b + 0 + 0}{3} = \frac{b}{3}$.This gives $a = 3h$ and $b = 3k$.Substituting these into the circle equation $a^2 + b^2 = 9$,we get $(3h)^2 + (3k)^2 = 9$.$9h^2 + 9k^2 = 9$,which simplifies to $h^2 + k^2 = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 1$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

$B$ and $C$ are fixed points having coordinates $(3, 0)$ and $(-3, 0)$ respectively. If the vertical angle $\angle BAC$ is $90^o$,then the locus of the centroid of the $\Delta ABC$ has the equation:

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