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Find the inverse point of (-2,3) with respect to the circle x^2+y^2-4x-6y+9=0 Find the inverse point of (-2,3) with respect to the circle x^2+y^2-4x-6y+9=0
The equation of the circle isx²+y²-4x-6y+9=0This can be written as(x-2)²+(y-3)²=4=2² so centre of circle is O (2 3) and radius r =2Now let a point P(h,k ) is a inverse of poin A (-2,3) w.r.t. circle x²+y²-4x-6y+9=0 So OP x OA=r²=4 Now OA=√{ (2+2)²+(3-3)²}=√4²=4Thus OP x 4=4OP=1OP²=1......................(1) Now equation of OA is:y-3=(3-3)/(-2-2) *(x-2)y-3=0y=3P (h,k) lies on OPso k=3From(1) OP²=1(h-2)²+(k-3)²=1(h-2)²+(3-3)²=1(h-2)²=1h-2=±1h=3 or 1Thus the points are (1,3 ) or (3,3)But point P is same side of point A that is possible only(1,3)Thus point P is (1,3)
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