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
formula of a circle is $\displaystyle (x-h)^2+(y-k)^2 = r^2$
by observation h=5 and k=4 and R=5
so
$\displaystyle (x-5)^2+(y-4)^2 = 25$
at least if I am understanding this correctly..
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
General equation of circle: $\displaystyle x^2+y^2+2gx+2fy+c=0$
The circle is tangent to the y-axis(x=0) implies the discriminant of the intersection equals 0.
$\displaystyle y^2+2fy+c=0$
$\displaystyle (2f)^2-4c=0 $---1
Then plug in (2,0) and (8,0) into the general equation and solve the 3 simultaneous equations.
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.
$\displaystyle (x-a)^2 + (y-b)^2 = r^2$
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