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 :
Show Hint
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 $2\sqrt{r^2-k^2} = 6\sqrt{5}$.
To solve the problem, we need to determine the radius of the circle \( C \) given its properties.
Let's identify the key information from the question:
The circle touches the y-axis at the point \((0, 6)\).
It cuts off an intercept of length \(6\sqrt{5}\) on the x-axis.
Let's proceed to solve this using the circle's equation and properties:
Since the circle touches the y-axis at \((0, 6)\), the center of the circle can be assumed to be \((h, 6)\) where \(h\) is the radius of the circle because a circle touching the y-axis must have its center's x-coordinate equal to the radius.
The standard equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2
Here, \((h, k)\) is the center and \(r\) is the radius.
The circle cuts the x-axis at two points, where the length of the intercept is \(6\sqrt{5}\). The x-intercepts occur when \(y = 0\).
Substituting \(y = 0\) into the circle's equation:
(x - h)^2 + (0 - 6)^2 = h^2
Simplifying, we have:
(x - h)^2 + 36 = h^2 This implies:
(x - h)^2 = h^2 - 36
Since the circle cuts off an intercept of \(6\sqrt{5}\), the difference between the x-intercepts is \(6\sqrt{5}\), so:
2\sqrt{h^2 - 36} = 6\sqrt{5}
