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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.

Thank you very much!

May 10, 2020

#1
0

Didn't you post this problem already?

May 10, 2020
#2
0

This is my first time posting this question. If you could lead me to the original question that would be great!

May 10, 2020
#3
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May 10, 2020
#4
+21957
+1

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).)

May 10, 2020
#5
+31296
+3

I get a the same answer as Geno, obtained in a different way:

May 11, 2020
#6
+1421
+2

M is a midpoint of AB

Angles  MAC  and  ABO are equal

tan(ABO) = 3 / 12  = 14.03624347º

AM = {sqrt[(AO)² + (BO)²]} /2 = 6.184658438

r = AM / cos(MAC) = 6.375

May 11, 2020