- #1

Helly123

- 581

- 20

## Homework Statement

## Homework Equations

y-y1 = m (x-x1) ---> line equation

$$ (x - a)^2 + (y-b)^2 = r^2 $$ ---> circle equation

## The Attempt at a Solution

I tried to draw the triangles using, (1, 3) (2, 4) and (0, b)

(0, b) is the tangent point to y-axisand used those points for making perpendicular bisector, so I can find the center of the circle, and find the distance from that center to one of the points, which that distance is the radii.between (1, 3) (2, 4)

the mid point = 1.5 , 3.5 gradient = 1, the perpendicular m = -1

the line equation = y - 1.5 = -1(x - 3.5) ... (1)between (2,4) (0, b)

$$ mid point = (1, \frac {4+b)} {2} )

\\gradient = \frac {4-b}{2} \\the perpendicular gradient = \frac {-2}{4-b}

\\the line equation = y - \frac {(4+b)} {2} = \frac {-2}{(4-b)} (x - 1) ...(2) $$between (1,3) (0,b)

$$ \\mid point = (0.5, \frac {(3+b)} {2} )

\\gradient = \frac {(3-b)}{1} \\the perpendicular gradient = \frac {-1}{(3-b)}

\\the line equation = y - \frac {(3+b)} {2} = \frac {-1}{(3-b)} (x - 0.5) ...(3)

$$if I find the x, y then its the center

we know that the center (x, b) where b the same in b (0, b)

but i don't think its work.

can anyone give me another way to solve?

**I find another solutions but I don't get what the person doing can anyone explain it?**

https://www.algebra.com/algebra/homework/Circles/Circles.faq.question.1077722.html

https://www.algebra.com/algebra/homework/Circles/Circles.faq.question.1077722.html