Question

if a point P is moving such that the lengths of tangents drawn from P to the circles X square + Y square - 4 x -6 Y - 12 = 0 and X square + Y square + 6 X + 18 Y + 26 =0 are in the ratio 2:3,then find the equation to the locus of P​

Answer 1

$\textbf{Given:}$

$\text{Equation of circles are}$

$x^2+y^2-4x-6y-12=0$

$x^2+y^2+6x+18y+26=0$

$\textbf{To find:}$

$\text{The locus of P}$

$\textbf{Solution:}$

$\textbf{We know that,}$

$\textbf{Length of the tangent from the point}\bf\,P(x_1,y_1) \textbf{to the circle}\\\bf\,x^2+y^2+2gx+2fy+c=0\;is$

$\bf\,PT=\sqrt{{x_1}^2+{y_1}^2+2gx_1+2fy_1+c}$

$\text{Let the coordinates of P be (h,k)}$

$\textbf{Given geometrical condition:}$

$\dfrac{PT_1}{PT_2}=\dfrac{2}{3}$

$\dfrac{\sqrt{h^2+k^2-4h-6k-12}}{\sqrt{h^2+k^2+6h+18k+26}}=\dfrac{2}{3}$

$\text{Squaring on bothsides, we get}$

$\dfrac{h^2+k^2-4h-6k-12}{h^2+k^2+6h+18k+26}=\dfrac{4}{9}$

$9(h^2+k^2-4h-6k-12)=4(h^2+k^2+6h+18k+26)$

$9h^2+9k^2-36h-54k-108=4h^2+4k^2+24h+72k+104$

$\text{Rearranging terms, we get}$

$\implies\,5h^2+5k^2-60h-126k-212=0$

$\therefore\textbf{The locus of P is}$

$\boxed{\bf\,5x^2+5y^2-60x-126y-212=0}$

Find more:

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Answer 2

Answer:

5x^2 +5y^2-60x-126y-212=0