Question

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

Answer 1

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Answer 2

Answer:

Concept:

Let S = 0 be a circle with center C and radius r. two points P, Q are said to be inverse points with respect to S = 0 if. 2. If P, Q are a pair of inverse points with respect to a circle S = 0 then Q is called inverse point of P.

Step-by-step explanation:

Given:

x²+y²-4x+6y+9=0

P(-2,3)
Find:

Inverse point of P(-2,3)
Solution:

 
$Given \ point \ is (-2,3)={P}(x, y) \\\\Given \ equation \ of \ circle \ x^{2}+y^{2}-4 x+6 y+9=0 \\\\compare with x^{2}+y^{2}+2 g x+2 f y+c=0 \\\\ \begin{aligned} g &=-2 ;f=-3 \quad c=9 \end{aligned}$

   center (-g,-f) = (2,3)

$r=\sqrt{4+9-9}=\sqrt{4}=2 \\\\inverse point with respect to circle with center (h, k) is x^{\prime}=\alpha(x-h)+h ; \quad\\\\ y^{\prime}=\alpha(y-k)+k \quad P(x, y)=(-2,3) \\\\ \alpha=\frac{r^{2}}{(x-h)^{2}+(y-k)^{2}}$

 $\begin{aligned}&\alpha=\frac{4}{(-2-2)^{2}+(3-3)^{2}}=\frac{4}{16 }=\frac{1}{4} ; \quad y^{2}=\left(\frac{1}{4}\right)(3-3)+3 \\&x^{\prime}=\frac{1}{4}(-2-2)+2 ; \quad y^{\prime}=\frac{1}{4}(3-3)+3 \\&x^{\prime}= 1 ; \quad y^{\prime}=3 \\\end{aligned}$

$\therefore \text { inverse pant is }\left(x^{\prime}, y^{\prime}\right)=(1,3)$

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