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If circle S1 has radius R1 and circle S2 has radius R2 then prove that circles
[S1/R1 + S2/R2]and [S1/R1 - S2/R2] are orthogonal
Dear Student,
Let us take a general case of a circle.
Say the equation of the circle s, is
x2 + y2 + 2gx + 2fy + c = 0
where radius R1 = (g2 + f2 - c)1/2
and say S2 is
x2 + y2 + 2g1x + 2f1y + c1 = 0
where radius R2 = (g12 + f12 - c1)1/2
Now in order to prove that the curves (S1/R1 + S2/R2 )
and (S1/R1 + S2/R2 ) cut each other orthogonally we have to show that dy/dx to the first curve at the point of intersection and dy/dx to the second curve at the same point when multiplied = –1
i.e., (dy/dx)c1 * (dy/dx)c2 = –1
Note : Students are advised to remember this property as a result.
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