# Write the equation of the circle? The points A(1;-2) and B(-1;2) are located on the circle and a line "d: 3x+4y-5=0" passes through the center of the circle.

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### 1 Answer

Let C(x,y) represent the center of the circle. Then the distance from A to C equals the distance from B to C (all radii are congruent.)

We use the distance formula : `d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)`

AC=`sqrt((x-1)^2+(y+2)^2)` and BC=`sqrt((x+1)^2+(y-2)^2)` AC=BC==>

`sqrt((x-1)^2+(y+2)^2)=sqrt((x+1)^2+(y-2)^2)` squaring both sides:

`(x-1)^2+(y+2)^2=(x+1)^2+(y-2)^2` Expanding:

`x^2-2x+1+y^2+4y+4=x^2+2x+1+y^2-4y+4` Collect like terms

`4x=8y`

`x=2y`

Now we notice that B lies on the line through the center; for the line 3x+4y-5=0 if x=-1 then 3(-1)+4y-5=0==>4y=8==>y=2 so the point (-1,2) which is B lies on that line. Since the center lies on that line, we can solve for y and substitute:

3x+4y-5=0==>`y=-3/4x+5/4` Substitute into x=2y:

`x=2(-3/4x+5/4)`

`x=-3/2x+5/2`

`5/2x=5/2==>x=1` so `y=-3/4(1)+5/4=1/2`

The center of the circle is at `(1,1/2)` .

To find the radius we can find either CA or CB. Note that CA is a vertical line so CA=2.5 is the radius.

Then we substitute into the equation of a circle: `(x-h)^2+(y-k)^2=r^2` where the center is (h,k) and the radius r.

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The equation of the circle is `(x-1)^2+(y-1/2)^2=6.25`

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The graph: