From a point $\mathrm{A}(0,3)$ on the circle $(x+2)^2$ $+(y-3)^2=4$, a chord $A B$ is drawn and it is…

From a point $\mathrm{A}(0,3)$ on the circle $(x+2)^2$ $+(y-3)^2=4$, a chord $A B$ is drawn and it is extended to a point $Q$ such that $A Q=2 A B$. Then the locus of $Q$ is
  1. $(x+4)^2+(y-3)^2=16$
  2. $(x+1)^2+(y-3)^2=32$
  3. $(x+1)^2+(y-3)^2=4$
  4. $(x+1)^2+(y-3)^2=1$

Solution

Given equation of circle $(x+2)^2+(y-3)^2=4$ Let the coordinates of Q is $(h, k)$. Coordinate of B which is midpoint of AQ because $\mathrm{AQ}=2 \mathrm{AB}$. Then, $\mathrm{B}=\left(\frac{0+h}{2}, \frac{k+3}{2}\right) \rightarrow\left(\frac{h}{2}, \frac{k+3}{2}\right)$ Point B also satisfy the equation of circle. $(x+2)^2+(y-3)^2=4$ $\left(\frac{h}{2}+2\right)^2+\left(\frac{k+3}{2}-3\right)^2=4$ $\frac{(h+4)^2}{4}+\frac{(k-3)^2}{4}=4$ $(h+4)^2+(k-3)^2=16$ Replace $(h, k)$ by $(x, y)$, then, the required equation is $(x+4)^2+(y-3)^2=16$

Asked in: BITSAT 2024 (Memory Based Paper 1)

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