1. ## coordinate geometry

two circles each of radius 5 units touch each other at (1,2) if the equation of common tangent is 4x+3y=10 find the equation of two circles

2. Do they touch each other anywhere else?

3. You will need to find the coordinate of the centre of both circles.

(1) Find the gradient of the line perpendicular to the tangent. Set up an equation.

(2) Given that the radius is 5, use the distance formula to set up another equation.

(3) Solve the system and decide which is the centre for the first and second circle.

4. Originally Posted by Prove It
Do they touch each other anywhere else?
Hmm...
common tangent is 4x+3y=10

5. Originally Posted by prasum
two circles each of radius 5 units touch each other at (1,2) if the equation of common tangent is 4x+3y=10 find the equation of two circles

The better way I think is by looking on triangle with basis 10 and bisection equation 4x+3y=10 .

6. Hello, prasum!

Here's a back-door approach . . .

Two circles each of radius 5 units touch each other at (1,2).
If the equation of common tangent is $4x+3y\,=\,10$,
find the equation of two circles.
Code:
         *
|\               * * *
| \          *           *
-4 |  \       *               *
|   \     *                 *
*- - *
3  \  *         P         *
\ *         ♠         *
\*       o |         *
* * *   \     o   |
*           **  o     |        *
*       * - - - ◊ - - - *       *
*        |     o  **           *
|   o     \   * * *
*         | o       *\
*         ♣         * \
*         Q         *  \
\
*                 *     \
*               *       \
*           *
* * *

The two circles are tangent at: . $\diamondsuit(1,2)$

The centers of the circles are: . $P\spadesuit \text{ and }Q\clubsuit$

The tangent has slope $-\frac{4}{3}$

The radii, $P\diamondsuit \text{ and }Q\diamondsuit$ are perpendicular to the tangent
. . and have length 5.

From $\diamondsuit(1,2)$ move 4 units right and 3 units up to $P(5,5)$
. . and we find that: . $|P\diamondsuit| = 5$

From $\diamondsuit(1,2)$ move 4 units left and 3 units down to $Q(\text{-}3,\text{-}1)$
. . and we find that: . $|Q\diamondsuit| = 5$

Hence, the centers of the circles are: . $P(5,5) \text{ and }Q(\text{-}3,\text{-}1)$

Therefore, their equations are: . $\begin{Bmatrix}(x-5)^2 + (y-5)^2 &=^& 25 \\ (x+3)^2 + (y+1)^2 &=& 25 \end{Bmatrix}$