1. cGeometry of circle

If I had the points p and q on the circumference of a circle such that p was 8,0 and q was 2,0 how would I find the equation if the y axis is tangential?
thankyou

2. formula of a circle is $(x-h)^2+(y-k)^2 = r^2$

by observation h=5 and k=4 and R=5

so

$(x-5)^2+(y-4)^2 = 25$

at least if I am understanding this correctly..

3. Are you saying there exists only one circle that can go through these points?

4. well as I understand this, a 3rd point is determined since the y axis is tangent to the circle. which makes the radius, H and K observable... my most accurate wild guess on the matter...
so 3 points determine a circle (and only one circle)
i had the same thots tho

5. Originally Posted by pickslides
Are you saying there exists only one circle that can go through these points?
No, there are 2

6. Originally Posted by bigwave
formula of a circle is $(x-h)^2+(y-k)^2 = r^2$

by observation h=5 and k=5 and R=5

so

$(x-5)^2+(y-5)^2 = 25$

at least if I am understanding this correctly..
This doesnt seem right though

7. $(x-a)^2 + (y-b)^2=r^2$

plug p=8.0 and q=2.0 and y=0

8. Originally Posted by p0oint
$(x-a)^2 + (y-b)^2=r^2$

plug p=8.0 and q=2.0 and y=0
thanks.....how would that work though?

9. Are p and q the points of the circle on the x-axis?

I mean what are the (x,y) of p and q?

10. the points are 2,0 and 8,0 are on the circle with the y axis being a tangent to the circle

11. Originally Posted by 200001
the points are 2,0 and 8,0 are on the circle with the y axis being a tangent to the circle
General equation of circle: $x^2+y^2+2gx+2fy+c=0$

The circle is tangent to the y-axis(x=0) implies the discriminant of the intersection equals 0.

$y^2+2fy+c=0$

$(2f)^2-4c=0$---1

Then plug in (2,0) and (8,0) into the general equation and solve the 3 simultaneous equations.

12. Hello 2oughts1
I read your problem as the point on the circle is (8,2). I see only one circle tangent to y axis from this point.Its center is (4,2). So the equation is now easily written.

bjh

13. Hello 2oughts1,
I read your problem as (8,2) is a point on the circle. I see only one circle tangent to y axis from this point and its center is (4,2). Take it from there.

bjh

14. Originally Posted by bjhopper
Hello 2oughts1,
I read your problem as (8,2) is a point on the circle. I see only one circle tangent to y axis from this point and its center is (4,2). Take it from there.

bjh
I don't want to pick at you but ... see attachment.

15. Nice graph earboth.

If I take the initial problem, and you want to solve it graphically, you know that the centre of the circle is somewhere on the line x = 5. then, take a compass, measure 5 units. The centre of the circle should be equidistant from both points. So, cut the line x = 5, from the points (2, 0) and (8, 0). Both should cut the line at the same point(s). This is the centre of your circle. Now, you'll find that there are two possible circles. One at a certain (x1,y1) and another at another (x2,y2).

Then, use the cartesian equation of a circle.

$(x-a)^2 + (y-b)^2 = r^2$

Replace r by the radius of the circle,
a by one of the x coordinates and b by the corresponding y coordinates.

Tell us what you get

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