The radius of the circle whose center lies at (1 ,2 ) while cutting the circle x^2+y^2+4x+16y-30=0 orthogonally, is .......... units.

Question

Accepted Answer

Correct answer: D. (d) Given, circle x^2+y^2+4x+16y-30=0 … . (i ) Where, g_1=2 ,f_1=8 ,c_1=-30 Now equation of circle whose centre at (1 ,2 ) is x^2+y^2-2x-4y+c=0 Here, g_2=-1 ,f_2=-2 ,c_2=c Since, circles cut each other orthogonally So , 2(g_1g_2+f_1f_2)=c_1+c_2 ⇒2 (-2 ,-16 )=-30+c or c=-6 From eqs. (ii), we get x^2+y^2-2x-4y-6=0 Radius r= (-1 )^2+ (-2 )^2- (-6 )= 11

$The radius of the circle whose center lies at (1, 2) while cutting the circle x^{2} + y^{2} + 4 x + 16 y - 30 = 0 orthogonally, is .......... units.$

Solution

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