1. ## Circle

What is the equation of the circle passing through (12,1) and (2,-5) with center on the line 2x-5y+10=0.

2. Originally Posted by lance
What is the equation of the circle passing through (12,1) and (2,-5) with center on the line 2x-5y+10=0.

using the standard equation of a circle:x^2+y^2+2gx+2fy+c=0 we get 2 eqns in f, g and c. since centre is given by (-g,-f) it lies on the given line.
you have 3 eqns for 3 variables. so you should get a solution.

3. Let $\displaystyle A(12,1), \ B(2,-5)$.

Let $\displaystyle C(a,b)$ be the center of the circle.

Let $\displaystyle d: \ 2x-5y+10=0$.

Then $\displaystyle C\in d\Rightarrow 2a-5b+10=0$

$\displaystyle CA=CB\Leftrightarrow CA^2=CB^2\Leftrightarrow (a-12)^2+(b-1)^2=(a-2)^2+(b+5)^2$.

Now you have to solve the system

$\displaystyle \left\{\begin{array}{ll}2a-5b+10=0\\(a-12)^2+(b-1)^2=(a-2)^2+(b+5)^2\end{array}\right.$

Then find $\displaystyle r=CA$

The equation of the circle will be $\displaystyle (x-a)^2+(y-b)^2=r^2$.

4. ## circle

postes by Lance,

Hi Lance,
Equation of circle centered @ h,k where h is x coordinate and k is y coordinate on x-y graph

(x-h)^2 +(y-k)^2=r^2

Plot ihe two points and the given line on graph paper.Connect the two points and find the midpoint of the resulting line using its slope diagram. Erect a perpendicular at the midpoint by taking the negative inverse of that slope.Draw the line with this new slope from the midpoint crossing the given line.The intersection is the center of the circle.(h,k). From this point draw aline to either given point.Erect a new slope diagram and calculate the radius.Plug values into the above circle equation.

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