Question

A circle with center (2, 2) touches the coordinate axes and a straight line AB where A and B lie on positive direction of coordinate axes such that the circle lies between origin and the line AB. If O be the origin then the locus of circumcenter of DOAB will be:

• A. (x-2)(y-2)=2
• B. (x-1)(y-1)=2
• C. (x-4)(y-4)=8
• D. (x-2)(y-2)=4

Answer 1

Answer: (A)

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

Answer 2

Let the equation of the line AB be $\frac{x}{a} + \frac{y}{b} = 1$, where $a > 0$ and $b > 0$. The circle with center (2, 2) touches the coordinate axes, so its radius is $r=2$. The distance from the center (2, 2) to the line AB must be equal to the radius 2. The equation of the line AB can be written as $bx + ay - ab = 0$. The distance formula gives: $\frac{|b(2) + a(2) - ab|}{\sqrt{b^2 + a^2}} = 2$ $|2a + 2b - ab| = 2\sqrt{a^2 + b^2}$ Squaring both sides: $(2a + 2b - ab)^2 = 4(a^2 + b^2)$ $4a^2 + 4b^2 + a^2b^2 + 8ab - 4a^2b - 4ab^2 = 4a^2 + 4b^2$ $a^2b^2 + 8ab - 4a^2b - 4ab^2 = 0$ Since $a > 0$ and $b > 0$, we can divide by $ab$: $ab + 8 - 4a - 4b = 0$ $ab - 4a - 4b + 16 = 8$ $(a-4)(b-4) = 8$

The circle lies between the origin and the line AB. This means the origin (0,0) and the center of the circle (2,2) are on the same side of the line AB. For the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$: At (0,0): $\frac{0}{a} + \frac{0}{b} - 1 = -1$. At (2,2): $\frac{2}{a} + \frac{2}{b} - 1$. For them to be on the same side, $\frac{2}{a} + \frac{2}{b} - 1 < 0$, which means $\frac{2}{a} + \frac{2}{b} < 1$. $\frac{2b + 2a}{ab} < 1$ $2a + 2b < ab$ $ab - 2a - 2b > 0$ $ab - 2a - 2b + 4 > 4$ $(a-2)(b-2) > 4$

Assuming DOAB is a rectangle, D has coordinates $(a, b)$. The circumcenter of rectangle DOAB (with vertices O(0,0), A(a,0), B(0,b), D(a,b)) is the midpoint of the diagonal OD, which is $\left(\frac{a}{2}, \frac{b}{2}\right)$. Let the circumcenter be $(h, k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$, so $a = 2h$ and $b = 2k$.

Substitute $a=2h$ and $b=2k$ into $(a-4)(b-4) = 8$: $(2h-4)(2k-4) = 8$ $4(h-2)(k-2) = 8$ $(h-2)(k-2) = 2$

Now, check the condition $(a-2)(b-2) > 4$: $(2h-2)(2k-2) > 4$ $4(h-1)(k-1) > 4$ $(h-1)(k-1) > 1$ Let $X = h-2$ and $Y = k-2$. Then $XY = 2$. The condition becomes $(X+1)(Y+1) > 1$. $XY + X + Y + 1 > 1$. $2 + X + Y + 1 > 1$. $X + Y + 3 > 1$. $X + Y > -2$. If $X, Y > 0$, then $h>2, k>2$. $X+Y > 0$, so $X+Y > -2$ is satisfied. If $X, Y < 0$, then $h<2, k<2$. Let $X = -u, Y = -v$ with $u, v > 0$ and $uv=2$. $X+Y = -(u+v)$. By AM-GM, $u+v \ge 2\sqrt{uv} = 2\sqrt{2}$. So $X+Y \le -2\sqrt{2}$. Since $-2\sqrt{2} \approx -2.828$, $X+Y < -2$. Thus, the condition $X+Y > -2$ is not satisfied when $X, Y < 0$. Therefore, the locus is $(h-2)(k-2) = 2$ with $h>2$ and $k>2$. The locus equation is $(x-2)(y-2)=2$.