If z = x + iy and the point P in the Argand d | vedclass

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(C) Given the equation $2|z - (2 + 3i)| = 3|z - (2 - i)|$. Let $z = x + iy$. Then $z - (2 + 3i) = (x - 2) + i(y - 3)$ and $z - (2 - i) = (x - 2) + i(y

Vedclass · Accepted Answer

$\left(2, -\frac{21}{5}\right)$

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(C) Given the equation $2|z - (2 + 3i)| = 3|z - (2 - i)|$. Let $z = x + iy$. Then $z - (2 + 3i) = (x - 2) + i(y - 3)$ and $z - (2 - i) = (x - 2) + i(y + 1)$. Squaring both sides,we get $4|z - (2 + 3i)|^2 = 9|z - (2 - i)|^2$. $4((x - 2)^2 + (y - 3)^2) = 9((x - 2)^2 + (y + 1)^2)$. $4(x^2 - 4x + 4 + y^2 - 6y + 9) = 9(x^2 - 4x + 4 + y^2 + 2y + 1)$. $4x^2 - 16x + 16 + 4y^2 - 24y + 36 = 9x^2 - 36x + 36 + 9y^2 + 18y + 9$. Rearranging terms to one side: $5x^2 + 5y^2 - 20x + 42y + 3 = 0$. Dividing by $5$: $x^2 + y^2 - 4x + \frac{42}{5}y + \frac{3}{5} = 0$. The centre of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $(-g, -f)$. Here,$2g = -4 \implies g = -2$ and $2f = \frac{42}{5} \implies f = \frac{21}{5}$. Thus,the centre is $(2, -\frac{21}{5})$.

If $z = x + iy$ and the point $P$ in the Argand diagram represents $z$,then the locus of the point $P$ satisfying the equation $2|z - 2 - 3i| = 3|z - 2 + i|$ is a circle with centre

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