1. ## coordinate geometry

find the locus of centres of circle x^2+y^2-2ax-2by+2=0 where a and b are parameters if the tangents from any to each of the circles are orthogonal

2. Hello, prasum!

I'm not sure I understand the problem . . .

Find the locus of centres of circle: $\displaystyle x^2+y^2-2ax-2by+2\:=\:0$
. . where $\displaystyle a$ and $\displaystyle b$ are parameters,
if the tangents from any point to each of the circles are orthogonal.

The circle has the equation: .$\displaystyle (x-a)^2 + (y-b)^2 \:=\:a^2+b^2-2$

It has center $\displaystyle C(a,b)$ and radius $\displaystyle r\,=\,\sqrt{a^2+b^2-2}$

Let $\displaystyle P(h,k)$ be any point (exterior to the circle).

Tangents are drawn from $\displaystyle P$ to $\displaystyle A$ and $\displaystyle B$ on the circle.
. . The tangents are orthogonal: .$\displaystyle \angle P = 90^o$

Code:
    |
|         * * *
|     *           * A
|   *               ♥
|  *              o  *o
|               o       o
| *       C   o       *   o    P
| *    (a,b)♥         *     ♥(h,k)
| *           o       *   o
|               o       o
|  *              o  *o
|   *               ♥
|     *           * B
|         * * *
- - + - - - - - - - - - - - - - - - - - - - - -
|

Since $\displaystyle \angle A = \angle B = \angle P = 90^o,\;PA = PB,\:CA = CB$
. . then quadrilateral $\displaystyle APBC$ is a square.

Its diagonal is: .$\displaystyle CP \,=\,\sqrt{2}\,r \;=\;\sqrt{2(a^2+b^2-2)}$

Therefore, the locus of the centers of the circles
. . is a circle with center $\displaystyle (h,k)$ and radius $\displaystyle \sqrt{2(a^2+b^2-2)}$