Question

Two circles with equal radii are intersecting at the points (0, 1) and (0, -1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

• A. 1
• B. √2
• C. 2√2
• D. 2

Answer 1

Answer: (D)

2

Answer 2

Let the two circles be $C_A$ and $C_B$, with centers $C_A$ and $C_B$ respectively, and let their common radius be $r$. The circles intersect at $A=(0, 1)$ and $B=(0, -1)$. The line segment $AB$ is the common chord, and its midpoint is $(0,0)$. The centers $C_A$ and $C_B$ must lie on the perpendicular bisector of $AB$, which is the x-axis. Let $C_A = (h, 0)$ and $C_B = (-h, 0)$ for some $h > 0$. The distance between the centers is $2h$. Since $A(0, 1)$ lies on circle $C_A$, the radius $r$ is the distance $C_A A$: $r^2 = (0-h)^2 + (1-0)^2 = h^2 + 1$. The tangent to circle $C_A$ at $A(0, 1)$ passes through $C_B(-h, 0)$. The radius $C_A A$ is perpendicular to the tangent $A C_B$. The vector $C_A A = (-h, 1)$ and the vector $A C_B = (-h, -1)$. Their dot product is zero: $(-h)(-h) + (1)(-1) = h^2 - 1 = 0$. Thus, $h^2 = 1$, and since $h > 0$, $h=1$. The distance between the centers is $2h = 2(1) = 2$.