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 $(4,2)$.

recepiamsb

Beginner2022-09-06Added 9 answers

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 $(1/2)((0,0)+(4,2))=(2,1)$. 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

$y-1=-2(x-2),y=-2x+5$

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 $(4,-3)$. The radius should now be easy to figure out given that $(4,2)$ is on the circle centered at $(4,-3)$, and the equation of the circle follows.

$y-1=-2(x-2),y=-2x+5$

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 $(4,-3)$. The radius should now be easy to figure out given that $(4,2)$ is on the circle centered at $(4,-3)$, and the equation of the circle follows.

Rohan Mcpherson

Beginner2022-09-07Added 1 answers

Hint

Let the center be $C(c,1-c)$

If $r$ is the radius,

${r}^{2}={c}^{2}+(1-c{)}^{2}=(c-4{)}^{2}+(1-c-2{)}^{2}$

The last equation will give us the value of $c$

Let the center be $C(c,1-c)$

If $r$ is the radius,

${r}^{2}={c}^{2}+(1-c{)}^{2}=(c-4{)}^{2}+(1-c-2{)}^{2}$

The last equation will give us the value of $c$

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