Question

The equation of the circle whose two diameters are the lines $x+y=4$ and $x-y=2$ and which passes through $(4,6)$ is

• A. $x^2+y^2-6x-2y-16=0$
• B. $x^2+y^2-6x-2y=15$
• C. $x^2+y^2=0$
• D. $5(x^2+y^2)-4x=16$

Answer 1

The correct option is A $x^2+y^2-6x-2y-16=0$

Two diameters are the lines,

$x+y=4$ ----- ( 1 )

$x-y=2$ ----- ( 2 )

So, first, we find the intersection point of diameters that is the center of the circle.

Adding equation ( 1 ) and ( 2 ) we get,

$\Rightarrow$ $2x=6$

$\Rightarrow$ $x=3$

Substituting value of $x$ in equation ( 1 ),

$\Rightarrow$ $3+y=4$

$\Rightarrow$ $y=1$

So, the center of circle is $(3,1).$

Now, circle is pass through $(4,6)$

The equation of circle is $(x-a{)}^2+(y-b)=r^2$, where $(a,b)$ co-ordinates of circle and $r$ is radius.

Put center coordinates and pass through point

$\Rightarrow$ $(4-3{)}^2+(6-1{)}^2=r^2$

$\Rightarrow$ $1+25=r^2$

$\Rightarrow$ $r^2=26$

So, the equation of circle is,

$\Rightarrow$ $(x-3{)}^2+(y-1{)}^2=26$

$\Rightarrow$ $x^2-6x+9+y^2-2y+1=26$

$\Rightarrow$ $x^2+y^2-6x-2y-16=0$