1. ## Equation of circle

Find the equation of the circle with passes through the points $(0,1)$ and $(3,-2)$ and has its centre lying on the line $y=x-2$

2. Originally Posted by Punch
Find the equation of the circle with passes through the points $(0,1)$ and $(3,-2)$ and has its centre lying on the line $y=x-2$
If the point $(a,b)$ is the center then

$r=\sqrt{(0-a)^2+(1-b)^2} \iff r^2=(0-a)^2+(1-b)^2$ and $r=\sqrt{(3-a)^2+(-2-b)^2} \iff r^2=(3-a)^2+(-2-b)^2$ and since the point lies on the line we get

$b=a-2$

Now set the $r^2$ equations equal to each other and sub in $b=a-2$ This will give you a quadratic equation in a.

3. Originally Posted by Punch
Find the equation of the circle with passes through the points $(0,1)$ and $(3,-2)$ and has its centre lying on the line $y=x-2$
Similar to your other post, there is also a problem with this!

The radius and centre cannot be determined, since an infinite number of circles pass through those two points
and have their centres on the given line.

Are there any other details to the question ??

4. Originally Posted by Archie Meade
Similar to your other post, there is also a problem with this!

The radius and centre cannot be determined, since an infinite number of circles pass through those two points
and have their centres on the given line.

Are there any other details to the question ??
No, I have checked and there are no other details

5. If the points (0,1) and (3,-2) are the endpoints of the diameter of the circle,
then the centre is (1.5,-0.5), the radius is obtained from Pythagoras' theorem.

The equation of the circle then is $(x-1.5)^2+(y+0.5)^2=r^2$

However, if the line from (0,1) to (3,-2) is a chord of the circle,
then there are an infinite number of solutions,
since you can choose any point on the line y=x-2 as the circle centre.

The line is perpendicular to the line joining the two given points,
and passes through the midpoint of the chord.

6. Originally Posted by Archie Meade
If the points (0,1) and (3,-2) are the endpoints of the diameter of the circle,
then the centre is (1.5,-0.5), the radius is obtained from Pythagoras' theorem.

The equation of the circle then is $(x-1.5)^2+(y+0.5)^2=r^2$

However, if the line from (0,1) to (3,-2) is a chord of the circle,
then there are an infinite number of solutions,
since you can choose any point on the line y=x-2 as the circle centre.

The line is perpendicular to the line joining the two given points,
and passes through the midpoint of the chord.
In conclusion, there is a problem with the question. Am I right?

7. Originally Posted by Punch
In conclusion, there is a problem with the question. Am I right?
Hi Punch,

yes, the question is incomplete.
You can certainly find the equation of a circle, as you can pick any point on y=x-2 as the centre of the circle.

However, if you want the equation of the circle, then a further condition must be provided with the question.