flatantsmu

2022-09-05

Find the equation of the following circles: A circle has its centre on the line $x+y=1$ and passes through the origin and the point $\left(4,2\right)$.

recepiamsb

Given any two points on a circle the center lies on to the perpendicular bisector of the chord between them. Here the perpendicular bisector passes through the midpoint $\left(1/2\right)\left(\left(0,0\right)+\left(4,2\right)\right)=\left(2,1\right)$. The slope of the chord is clearly $+1/2$ so, by the "negative reciprocal rule" the perpendicular line has slope $-2$. So the center lies on
$y-1=-2\left(x-2\right),y=-2x+5$
Since the center also lies on $y=1-x$ we then have for the center:
$1-x=-2x+5,x=4,y=-3$
meaning the center is $\left(4,-3\right)$. The radius should now be easy to figure out given that $\left(4,2\right)$ is on the circle centered at $\left(4,-3\right)$, and the equation of the circle follows.

Rohan Mcpherson

Hint
Let the center be $C\left(c,1-c\right)$
If $r$ is the radius,
${r}^{2}={c}^{2}+\left(1-c{\right)}^{2}=\left(c-4{\right)}^{2}+\left(1-c-2{\right)}^{2}$
The last equation will give us the value of $c$

Do you have a similar question?