# Math Help - analytic geometry in a plane - cirle

1. ## analytic geometry in a plane - cirle

find the equation of the circle that passes through the point ( 9,7) and is tangent to both the y-axis and line 3x-4y-24=0.

What would be the best way to answer this question? can you please give me an illustration?
I'm just really stuck on this one.

2. Originally Posted by jam2011
find the equation of the circle that passes through the point ( 9,7) and is tangent to both the y-axis and line 3x-4y-24=0.

What would be the best way to answer this question? can you please give me an illustration?
I'm just really stuck on this one.
1. Draw a sketch!

2. The equation of the circle is

$(x-x_M)^2+(y-y_M)^2=r^2$

3. $x_M=r$ Why?

4. The point P(9, 7) is located on the circle-line. That means:

$(9-x_M)^2+(7-y_M)^2=r^2$

5. The distance of the midpoint M of the circle to the given line is r:

$\left| \dfrac{3x_M-4y_M-24}{\sqrt{3^2+4^2}} \right|=r$

6. Solve for $x_M, y_M, r$. There are two valid solutions.

3. ## interesting

i love to solve this but what formula can i use? hhhmmm ..tnx

4. Originally Posted by rcs
i love to solve this but what formula can i use? hhhmmm ..tnx
1. You can't use a single formula to solve the question but you have to solve a system of equations:

$\left|\begin{array}{rcl}(9-x_M)^2+(7-y_M)^2&=&r^2 \\ x_M&=&r \\ \left|\dfrac{3x_M-4y_M-24}5 \right|&=&r \end{array}\right.$

2. The 3rd equation has to be transformed into 2 different equations:

$\left|\dfrac{3x_M-4y_M-24}5 \right|=r~\implies~\left\{\begin{array}{l}\dfrac{3 x_M-4y_M-24}5 =r \\ \dfrac{3x_M-4y_M-24}5 =-r\end{array}\right.$

3. To shorten the calculations: Use the equation with the negative sign. Then you'll get:

$\left|\begin{array}{rcl}(9-x_M)^2+(7-y_M)^2&=&r^2 \\ x_M&=&r \\ \left|\dfrac{3x_M-4y_M-24}5 \right|&=&r \end{array}\right.~\implies~(9-x_M)^2+(7-(2x_M-6))^2=x_M^2$

4. This is a simple quadratic equation which I leave for you. Have fun!