# coordinate geometry

• Jul 20th 2010, 02:29 AM
prasum
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
• Jul 20th 2010, 02:00 PM
Soroban
Hello, prasum!

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

Quote:

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 $\displaystyle \sqrt{2(a^2+b^2-2)}$