Question

A circle touches the y-axis at $(0,2)$ and has an intercept of $4$ units on the positive side of the x-axis. Then the equation of the circle is?

• A. $x^2+y^2-4(\sqrt{2}x+y)+4=0$
• B. $x^2+y^2-4(x+\sqrt{2}y)+4=0$
• C. $x^2+y^2-2(\sqrt{2}x+y)+4=0$
• D. none of these

Answer 1

Answer: (A)

$x^2+y^2-4(\sqrt{2}x+y)+4=0$

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

$S(0,2)=0$

$⟹4+4f+c=0-eq.1$

Intercept on x-axis is given by

$2\sqrt{g^2-c}=4$

$g^2–c=4-eq.2$

Since x = 0 is a tangent

Radius = perpendicular distance from the center to tangent

$\sqrt {g^2 + f^2 - c} = \dfrac{|-g|}{\sqrt{1 + 0}$

$g^2+f^2-c=g^2$

$f^2=c$

Substituting in eq.1

$4+4f+f^2=0⟹(f+2{)}^2=0⟹f=-2$

$⟹c=4$

From eq.2

$g^2–4=4$

$g=\pm 2\sqrt{2}$

Equation of circle is

$x^2+y^2\pm 4\sqrt{2}x–4y+4=0$

$⟹x^2+y^2–4(\sqrt{2}x+y)+4=0$