Find the inverse point of P(-2,3) with the circle x²+y²-4x-6y+9=0

1.	Find the inverse point of P(-2,3) with the circle x²+y²-4x-6y+9=0
Answer» The inverse point of P(-2,3) with the circle x²+y²-4x-6y+9=0 is (1,3) Step-by-step explanation: The given equation of the circle is $x^2+y^2-4x-6y+9=0$ $\implies (x^2-2\times2x+4)+(y^2-2\times 3y+9)+9-4-9=0$ $\implies x^2-2\times x\times 2+2^2)+(y^2-2\times y]times 3+3^2)=4$ $\implies (x-2)^2+(y-3)^2=2^2$ Clearly, we can say that the centre of the circle is $(2,3)$ and radius is $2$ units We KNOW that inverse of a point P (x',y') w.r.t. a circle with centre at (h,k) and radius r is point (x",y") such that $x$ And, $y$ Where, $\alpha=\frac{r^2}{(x'-h)^2+(y'-k)^2}$ The given point is (-2,3) Thus, $\alpha=\frac{4}{(-2-2)^2+(3-3)^3}$ $\implies \alpha=\frac{1}{4}$ Therefore, the x-COORDINATE of the inverse point $x$ $\implies x$ The y-coordinate of the inverse point $x$ $\implies y$ Therefore, the coordinate of the inverse point is $(1,3)$ Hope this answer is helpful. Know More: Q: The inverse of (1,3) with the RESPECT of the circle x2+y2-4x-6y+9=0 Click Here: brainly.in/question/17419666

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