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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
```

4 years ago

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```										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
```
4 years ago
```										sorry,there was one mistake in above solution.
Ans. Is y^2=16x.
```
4 years ago
```										how?plz tell me..urgent plz..
```
4 years ago
```										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=0We 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-cBy equating them   f2-4g=c+4--(3)So (2)=(3) then f2=-16gLocus of (-x,-y) by sub y2=16x
```
5 days ago

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