Question

How do you write the equation of a circle centered at (-5,2) and contains the points (7,3)?

Answer 1

`(x+5)^2+(y-2)^2=145`

Explanation:

The formula of a circle centred at the origin is `x^2+y^2=r^2` because you are using the principle of Pythagoras on a triangle.

When the centre of the circle is not at the origin you mathematically move it back to centre.

So we have

`(x+5)^2+(y-2)^2=r^2`

Now we need to find the length `r` and this is achieved by calculating distance from centre to the given point using the well known
`a^2+b^2=c^2` for a triangle

`r=sqrt((-5-7)^2+(2-3)^2) = sqrt(12^2+1^2) =>r^2=145`

So we now have:

`(x+5)^2+(y-2)^2=145`
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`color(blue)("Comment:")`

`x_("centre")+5 = -5+5=0 larr" moved to origin"`
`y_("centre")-2 = +2-2=0 larr" moved to origin"`