Find the inverse point of (-2,3) with respect to the circle x^2+y^2-4x-6y+9=0

Question

Ram Kushwah · Answer

The equation of the circle is x²+y²-4x-6y+9=0 This can be written as (x-2)²+(y-3)²=4=2² so centre of circle is O (2 3) and radius r =2 Now 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=4 Thus OP x 4=4 OP=1 OP²=1......................(1) Now equation of OA is: y-3=(3-3)/(-2-2) *(x-2) y-3=0 y=3 P (h,k) lies on OP so k=3 From(1) OP²=1 (h-2)²+(k-3)²=1 (h-2)²+(3-3)²=1 (h-2)²=1 h-2= ±1 h=3 or 1 Thus 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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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

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