Question

# A circle of radius 5 units passes through the points (7,1),(9,5). If the ordinate of the centre is less than 2, then the equation of the circle is

A
x2+y2+8x10y+16=0
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B
x2+y2+8x+10y+16=0
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C
x2+y224x2y+120=0
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D
x2+y2+24x2y120=0
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Solution

## The correct option is D x2+y2−24x−2y+120=0Let (h,k) is a centre of circle , so(h−7)2+(k−1)2=25and (h−9)2+(k−5)2=25Letx2+y2+2gx+2fy+c=0 is circle .so, (7,1) lies on circle so, 50+14g+2f+c=0 ----(1)(9,5) lies on circle106+18g+10f+c=0 ----(2)and g2+f2−c=25 -----(3)From (1) & (2),4g+8f+56=0g+2f+14=0From (1), (2), (3), (4),g=−12 and f=−1,c=120so, eqn of circle isx2+y2−24x−2y+120=0

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