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find the equation of the circle whose centre is (3’-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0 the answer is given to be x^2+y^2+2x-4y-20=0 find the equation of the circle whose centre is (3’-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0 the answer is given to be x^2+y^2+2x-4y-20=0
If the circle cuts an intercept of 6 units of that line, that means length of the chord inside the circle is 6 units.now drop a perpendicular from center to the line, it should bisect the chord.Length of perpendicular dropped from center can be found out by p=|2*3-5*(-1)+18|/root(2^2+5^2) =root(29) From Pythagoras theorem,root(p^2+3^2)=r^2, where r is the radius of the circle=>r=root(38) hence, equation of circle is(x-3)^2 + (y+1)^2 = 38
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