Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $6\sqrt{5}$ on the x-axis. Then the radius of the circle C is equal to :
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The length of the intercept made by a circle $(x-h)^2 + (y-k)^2 = r^2$ on the x-axis is $2\sqrt{r^2-k^2}$ and on the y-axis is $2\sqrt{r^2-h^2}$. For this problem, the y-intercept is 0 (touches y-axis), so $r=|h|$. The x-intercept is $2\sqrt{r^2-k^2} = 6\sqrt{5}$.
Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.
The circle touches the y-axis at the point (0, 6).
Since it touches the y-axis, the distance from the center (h, k) to the y-axis is equal to the radius r. This distance is $|h|$. So, $r = |h|$.
Also, the point of tangency (0, 6) must lie on the circle. The center of the circle must have a y-coordinate equal to the y-coordinate of the point of tangency, so $k=6$.
The center of the circle is $(h, 6)$. Since $r = |h|$, let's assume the circle is in the first quadrant, so the center is $(r, 6)$.
The equation of the circle becomes $(x-r)^2 + (y-6)^2 = r^2$.
The circle cuts off an intercept on the x-axis of length $6\sqrt{5}$.
To find the x-intercepts, we set $y=0$ in the circle's equation.
$(x-r)^2 + (0-6)^2 = r^2$.
$(x-r)^2 + 36 = r^2$.
$x^2 - 2rx + r^2 + 36 = r^2$.
$x^2 - 2rx + 36 = 0$.
This is a quadratic equation for the x-coordinates of the intersection points, let them be $x_1$ and $x_2$.
The length of the intercept is $|x_2 - x_1|$.
We know that for a quadratic equation $ax^2+bx+c=0$, the difference between the roots is $|x_2 - x_1| = \frac{\sqrt{b^2 - 4ac}}{|a|}$.
Here, $a=1, b=-2r, c=36$.
Length of intercept = $\frac{\sqrt{(-2r)^2 - 4(1)(36)}}{1} = \sqrt{4r^2 - 144}$.
We are given that the length of the intercept is $6\sqrt{5}$.
$\sqrt{4r^2 - 144} = 6\sqrt{5}$.
Square both sides:
$4r^2 - 144 = (6\sqrt{5})^2 = 36 \times 5 = 180$.
$4r^2 = 180 + 144 = 324$.
$r^2 = \frac{324}{4} = 81$.
$r = \sqrt{81} = 9$.
The radius of the circle is 9.
