# Analytic Geometry

• Feb 18th 2010, 03:18 AM
lance
Analytic Geometry
Find the equation of the circle touching the x-axis and passing through the points (7,-2) and (0,-9).

• Feb 18th 2010, 06:50 AM
Pulock2009
Quote:

Originally Posted by lance
Find the equation of the circle touching the x-axis and passing through the points (7,-2) and (0,-9).

from the given points it can be seen that the circle touches both the axes. also the circle lies in the fourth quadrant.now since the axes are tangents the centre of the circle lies on y=-x. so the centre is (9,-9). you are given 2 points. so calculate the radius. then using circles equation get the required equation.
• Feb 18th 2010, 07:11 AM
Soroban
Hello, lance!

This is a messy one . . .

Quote:

Find the equation of the circle touching the $x$-axis
and passing through the points $A(7,-2)$ and $B(0,-9).$

Code:

      |       |        T     ---+------*-o-*------------       | Ao  (h,0)  *       |*              *       Bo                *       |       *|        C        *       *|        *        *       *|      (h,k)      *       |       *                *       |*              *       |  *          *       |      * * *       |

$\text{The center is }C(h,k)\text{, directly below the point of tangency, }T(h,0)$

$CA, CB\text{ and }CT\text{ are radii. }\;\text{Hence: }\,CA \,=\,CB \,=\,k$

We have: . $\begin{array}{ccccc}
CA^2 & =& (h-7)^2 + (k+2)^2 &=& k^2\\
CB^2 &=& h^2 + (k+9)^2 &=& k^2\end{array}$

Simplify: . $\begin{array}{ccccc}
h^2 - 14h + 4k + 53 &=& 0 & {\color{blue}[1]} \\
h^2 \qquad + 18k + 81 &=& 0 & {\color{blue}[2]} \end{array}$

Subtract [2] - [1]: . $14h + 14k + 28 \:=\:0 \quad\Rightarrow\quad k \:=\:-(h+2)\;\;{\color{blue}[3]}$

Substitute into [2]: . $h^2 - 18(h+2) + 81 \:=\:0 \quad\Rightarrow\quad h^2 - 18h + 45 \:=\:0$

Hence: . $(h-3)(h-15) \:=\:0 \quad\Rightarrow\quad h \:=\:3,\:15$

Substitute into [3]: . $k \:=\:-5,\:-17$

We have two solutions: . $\begin{array}{|c|c|ccc|} \hline
\text{Center} & \text{Radius} & \text{Equation} &&\\ \hline \\[-5mm]
(3,-5) & 5 & (x-3)^2 + (y+5)^2 &=& 25 \\ \\[-5mm]
(15,-17) & 17 & (x-15)^2 + (y+17)^2 &=& 289 \\ \hline\end{array}$

• Feb 18th 2010, 08:49 AM
lance
Hello!

Thank you very much Soroban... It really help me understand the problem... And also to you Pulock2009... (Happy)