Circles are drawn through the point (2, 0) to cut intercept of length 5 units on the x-axis. If their centers lie in the first quadrant, then find their equation.

Options

A. `x^(2)+y^(2)-9x+2fy+14=0`
B. `3x^(3)+3y^(2)+27x-2fy+42=0`
C. `x^(2)+y^(2)-9x-2fy+14=0`
D. `x^(2)+y^(2)-2fx-9y+14=0`

Answer

A

Explanation

the cricle `x^(2)+y^(2)+2gx+2fy+c=0`passes through(2,0).
`therefore" "4+4g+c=0`
its intercept on X-axis is `2xsqrt(g^(2)-c=5`
`rArr" "g^(2)-c=(25)/(4)rArrc=(4g^(2)-25)/(4)`
`therefore 4+4g+ (4+g^(2)-25)/(4)=0`
`rArr 16+16g+4g^(2)-25=0rArr4g^(2)+16g-9=0`
`rArr" "g=(-9)/(2),1/2` so,g=- 9/2, since centre lies in frist quadrant, also c=14. hence, `x^(2)+y^(2)-9x+2fy+14=0 ` is the required equatin.