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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 

Grade:11

1 Answers

Arun
25758 Points
4 years ago
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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