From a point A(0,3) on the circle (x+2)^2+(y- | vedclass

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Question

(A) Let the point $Q$ be $(h, k)$ and the point $B$ be $(x, y)$.Given that $AQ = 2AB$,this implies that $B$ is the midpoint of the segment $AQ$.Using

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$(x+4)^2+(y-3)^2=16$

Vedclass · Answer

(A) Let the point $Q$ be $(h, k)$ and the point $B$ be $(x, y)$.Given that $AQ = 2AB$,this implies that $B$ is the midpoint of the segment $AQ$.Using the midpoint formula,we have $x = \frac{h+0}{2} = \frac{h}{2}$ and $y = \frac{k+3}{2}$.The point $B(x, y)$ lies on the circle $(x+2)^2+(y-3)^2=4$.Substituting the values of $x$ and $y$ into the circle equation:$(\frac{h}{2}+2)^2 + (\frac{k+3}{2}-3)^2 = 4$$(\frac{h+4}{2})^2 + (\frac{k-3}{2})^2 = 4$$\frac{(h+4)^2}{4} + \frac{(k-3)^2}{4} = 4$$(h+4)^2 + (k-3)^2 = 16$Replacing $(h, k)$ with $(x, y)$,the locus of $Q$ is $(x+4)^2+(y-3)^2=16$.

From a point $A(0,3)$ on the circle $(x+2)^2+(y-3)^2=4$,a chord $AB$ is drawn and it is extended to a point $Q$ such that $AQ=2AB$. Then the locus of $Q$ is

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