Question

The equation of circle passing through the points $(1,\sqrt{2}),(7,\sqrt{2})and(1,3)$ is :-

• A. $x^2+y^2-8x-(3+\sqrt{2})y+7+3\sqrt{2}=0$
• B. $x^2+y^2+8x+(3+\sqrt{2})y+7+3\sqrt{2}=0$
• C. $x^2+y^2-8x-(3+\sqrt{2})y-7-3\sqrt{2}=0$
• D. $x^2+y^2+8x+(3+\sqrt{2})y-7-3\sqrt{2}=0$

Answer 1

Answer: (A)

$x^2+y^2-8x-(3+\sqrt{2})y+7+3\sqrt{2}=0$

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

Then as it passes through $(1,\sqrt{2})$

We get $2g+2\sqrt{2}f+c+3=0$ ...(1)

For $(7,\sqrt{2})$ we get $14g+2\sqrt{7}f+c+52=0$ ...(2)

And for $(1,3)$ we get $2g+6f+c+10=0$ ...(3)

Solving (1), (2) and (3) we get

$g=-4,f=\left(\frac{3+\sqrt{2}}{2}\right),c=7+3\sqrt{2}$