A circle passes through the point \((0,1),\) and is tangent to the parabola \(y=x^2\) at \((2,4).\) Find the center of the circle.
Any help would be greatly appreciated. Thank you!
+128460 The points (2,4) and ( 0,1) are on the circle
The slope of the tangent line to the parabola at (2,4) = 4
A line that perpendicular to this tangent line will have a slope of (-1/4)
And the equation of this line will be y = (-1/4)(x - 2) + 4 ⇒ (-1/4)x + 1/2 + 4 = (-1/4)x + 4.5 =
-x/4 + 4.5
And this line will pass through the center of the circle
Let m be the x coordinate of the center
Then y = -m/4 + 4.5 will be the y coordinate of the center
And this point will be equidistant from ( 0.1) and (2,4)
So using the distance formula we can set these distances equal and solve for m
(m - 0)^2 + (-m/4 + 4.5 - 1)^2 = (m -2)^2 + ( -m/4 + 4.5 - 4)^2 simplify
m^2 + ( 3.5 - m/4)^2 = m^2 - 4m + 4 + (.5 - m/4)^2
12.25 - (7/4)m + m^2/16 = -4m + 4 + .25 - m/4 + m^2/16
12.25 - (7/4)m = -4m + 4.25 -m/4
12.25 - 4.25 = (7/4 - 4 - 1/4) m
8 = -(5/2)m
8(-2/5) = m = -16/5 = -3.2 = x coordinate of the center
And the y coordinate of the center = -(1/4)(-16/5) + 4.5 = (16/20) + 4.5 = (4/5) + 4.5 = .8 + 4.5 = 5.3
So......the center = (-3.2, 5.3)
The radius^2 = (3.2)^2 + (5.3 -1)^2 = 28.73
The equation of the circle is
(x + 3.2)^2 + ( y - 5/3)^2 = 28.73
Here's a graph : https://www.desmos.com/calculator/4zun6t2nwj
