A circle passes through the point \((12,0)\), and is tangent to the \(y\)-axis at the point \((0,3),\) as shown. Find the radius of the circle.
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+23253 The tangent to the circle will be perpendicular to a diameter of the circle.
Therefore, the line y = 3 will contain the center of the circle.
Also, the center of the circle will be on the perpendicular bisector of each one of the chords.
Consider the chord whose endpoints are (0, 3) and (12, 0).
We need to find the equation of the perpendicular bisector of this chord.
Since the endpoints are (0, 3) and (12, 0), its midpoint is (6, 1.5).
The slope of this chord is: m = (0 - 3) / (12 - 0) = -3/12 = -1/4.
The slope of the perpendicular bisector of this chord is 4.
The equation of the perpendicular bisector is: y - 1.5 = 4(x - 6)
y - 1.5 = 4x - 24
y = 4x - 22.5
The intersection of this line with the line y = 3 is: 3 = 4x - 22.5
25.5 = 4x
x = 6.375
So, the center of the circle is (6.375, 3).
The radius will be 6.375. (This is the distance from the center to the point (0,3).)
