Question

How do you write the equation for a circle touching y-axis and passing though the points (1,5), (8,12)?

Answer 1

`(x-5)^2+(y-8)^2=25`

Explanation:

The general equation of a circle is
`(x-a)^2+(y-b)^2=r^2`
Where the centre is (a,b) and the radius is r
As the circle touches the y axis r=a
Draw any circle touching the y axis to see this.

So multiplying out gives:
`x ^2-2ax+a^2+y^2-2by+b^2=a^2`

This simplifies to
`x ^2-2ax+y^2-2by+b^2=0`

The point (1,5) is on the circle so substitute x =1 and y=5
Likewise (8,12) is on the circle.

You will have 2 simultaneous equation to find a and b.
Eliminate a to get to `56b=7b^2`

So b=8 or 0

Think about b= 0

The centre of the circle is on the y axis and the circle touches the y axis!!!

So b=8 and you can calculate the value of a