the locus of the center of the circle which cuts orthogonally the circle x2 +y2-20x +4=0 and which touches x=2 is

Question

Vaishal Shah · Answer

x2+y2+2gx+2fy+c=0 condition for orthogonal, 2*g*10=c+4 perpendicular fron center to x=2 is radius.. f2=22g put g=-x and f=-y hence,y2=-22x

Vaishal Shah · Answer

sorry,there was one mistake in above solution. Ans. Is y^2=16x.

KAMAL RAWAT · Answer

how?plz tell me..urgent plz..

Vincent · Answer

Let required circle be x2+y2+2gx+2fy=0---(1) Apply orthogonal condition (2gg'+2ff'=c+c') to above circle and to x2+y2-20x+4=0 We will get -20g=c+4---(2) Then apply perpendicular distance from centre (1) i.e (-g,-f) to give line x-2 Then we will get g+2 this is equal to radius of (1) i.e √g2+f2-c By equating them f2-4g=c+4--(3) So (2)=(3) then f2=-16g Locus of (-x,-y) by sub y2=16x

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the locus of the center of the circle which cuts orthogonally the circle x2 +y2-20x +4=0 and which touches x=2 is

the locus of the center of the circle which cuts orthogonally the circle x2 +y2-20x +4=0 and which touches x=2 is

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