The locus of a point which divides the join o | vedclass

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(B) Let the variable point on the circle be $B(x_1, y_1)$. Since $B$ lies on the circle ${x^2} + {y^2} = 4$,we have ${x_1}^2 + {y_1}^2 = 4$.Let the po

Vedclass · Accepted Answer

$25({x^2} + {y^2}) + 20(x - y) - 28 = 0$

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(B) Let the variable point on the circle be $B(x_1, y_1)$. Since $B$ lies on the circle ${x^2} + {y^2} = 4$,we have ${x_1}^2 + {y_1}^2 = 4$.Let the point dividing the segment $AB$ in the ratio $3 : 2$ be $P(h, k)$.Using the section formula,we have:$h = \frac{3x_1 + 2(-1)}{3 + 2} = \frac{3x_1 - 2}{5} \implies x_1 = \frac{5h + 2}{3}$$k = \frac{3y_1 + 2(1)}{3 + 2} = \frac{3y_1 + 2}{5} \implies y_1 = \frac{5k - 2}{3}$Substituting these values into the circle equation ${x_1}^2 + {y_1}^2 = 4$:$(\frac{5h + 2}{3})^2 + (\frac{5k - 2}{3})^2 = 4$$\frac{25h^2 + 20h + 4}{9} + \frac{25k^2 - 20k + 4}{9} = 4$$25(h^2 + k^2) + 20(h - k) + 8 = 36$$25(h^2 + k^2) + 20(h - k) - 28 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $25(x^2 + y^2) + 20(x - y) - 28 = 0$.

The locus of a point which divides the join of $A(-1, 1)$ and a variable point $B$ on the circle ${x^2} + {y^2} = 4$ in the ratio $3 : 2$ is:

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