Question

Two circles which touch both the coordinate axes intersect at the points A and B. If A = (1,2), then AB =

• A. 5
• B. 13
• C. $2\sqrt{2}$
• D. $\sqrt{2}$

Hint:

The radical axis of two circles $S_1=0$ and $S_2=0$ is the line $S_1-S_2=0$. This line contains the common chord of the two intersecting circles. For circles touching both axes, the centers lie on the line $y=x$, and their intersection points will be symmetric with respect to this line.

Answer 1

The Correct Option is D

A circle that touches both coordinate axes in the first quadrant has its center at $(r, r)$ and its equation is $(x-r)^2 + (y-r)^2 = r^2$.
Let the two circles be $C_1$ and $C_2$ with radii $r_1$ and $r_2$. Their equations are:
$S_1: (x-r_1)^2 + (y-r_1)^2 = r_1^2 \implies x^2 - 2r_1x + r_1^2 + y^2 - 2r_1y + r_1^2 = r_1^2 \implies x^2+y^2-2r_1x-2r_1y+r_1^2=0$.
$S_2: (x-r_2)^2 + (y-r_2)^2 = r_2^2 \implies x^2+y^2-2r_2x-2r_2y+r_2^2=0$.
The points of intersection A and B lie on both circles. The point A(1,2) lies on both circles, so its coordinates must satisfy both equations.
For the first circle: $1^2+2^2-2r_1(1)-2r_1(2)+r_1^2=0 \implies 5 - 2r_1 - 4r_1 + r_1^2 = 0 \implies r_1^2 - 6r_1 + 5 = 0$.
Factoring this gives $(r_1-1)(r_1-5)=0$. So the radii are $r_1=1$ and $r_2=5$.
The two circles are $x^2+y^2-2x-2y+1=0$ and $x^2+y^2-10x-10y+25=0$.
The line passing through the intersection points A and B is the radical axis, given by the equation $S_1 - S_2 = 0$.
$(x^2+y^2-2x-2y+1) - (x^2+y^2-10x-10y+25) = 0$.
$8x + 8y - 24 = 0 \implies x+y-3=0 \implies y=3-x$.
The two intersection points A and B are symmetric with respect to the line $y=x$. If A is (1,2), then B must be (2,1).
Let's verify that B(2,1) lies on the line $x+y-3=0$: $2+1-3=0$. Correct. Let's verify that B(2,1) lies on the first circle: $2^2+1^2-2(2)-2(1)+1 = 4+1-4-2+1=0$. Correct.
The coordinates of the intersection points are A(1,2) and B(2,1).
The distance AB is found using the distance formula:
$AB = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.