Question

The two diameters of a circle are segments of the straight lines $x-y=5$ and $2 x+y=4$. If the radius of the circle is 5 , then the equation of the circle is

• A. $x^2+y^2-6 x+4 y=12$
• B. $x^2+y^2-3 x+2 y=12$
• C. $x^2+y^2-6 x+2 y=12$
• D. $x^2+y^2-8 x+6 y-18=0$
• E. $x^2+y^2-8 x+6 y-7=0$

Answer 1

Answer: (A)

The intersecting point of lines $x-y=5$
and $2 x+y=4$ is $(3,-2)$
therefore Centre of circle is $(3,-2)$.
Required equation of circle is
$ (x-3)^2+(y+2)^2=(5)^2 $
$ \Rightarrow x^2+9-6 x+y^2+4+4 y=25 $
$ \Rightarrow x^2+y^2-6 x+4 y=12$