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`        Find the equation of circle whose Centre is( 3, - 1) and which cut off an intercept of length 6 from the line 2 x - 5y + 18 = 0`
8 months ago

```							hello ashutosh. first lets find the perpendicular distance of (3, – 1) from the line 2 x - 5y + 18 = 0.it would be p= |2*3 – 5*-1 + 18|/sqrt(2^2+5^2)= sqrt(29)now, we know that a perpendicular from (3, -1) to the line (which acts as a chord to the circle) would also bisect it. so by pythagoras theorem,r^2= p^2 + (6/2)^2 where r is radiusr^2= 29 + 9= 38so, eqn of circle:(x – 3)^2 + (y+1)^2= 38or x^2 + y^2 – 6x + 2y – 28= 0kindly approve :)
```
8 months ago
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