SOLUTION: find the equation of a circle that passes thru'(6,0) and (24,0) and is tangent to the y-axis

Question 993571: find the equation of a circle that passes thru'(6,0) and (24,0) and is tangent to the y-axis

Found 2 solutions by anand429, solver91311:
Answer by anand429(138) (Show Source): You can put this solution on YOUR website!
(6,0) and (24,0) both lie on x-axis, hence mid point of these points will lie directly below( or above) centre (Since line joining centre to mid point of a chord is perpendicular to the chord)
Also, since the circle touches the y axis, so distance from origin to this mid point found above will be equal to radius.
Now mid point is ((6+24)/2,(0+0)/2) i.e. (15,0)
So radius is 15.
Now let distance of centre from mid point found above be p
So, (DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)
i.e.
=> p = 12 or p = -12
So coordinates of centre will become (15,12) or (15,-12)
So, the equation of circle,
or
=> or
(DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

The segment of the x-axis bounded by the points (6,0) and (24,0) is a chord of the desired circle. Since the perpendicular bisector of a chord must pass through the center of the circle, we know that the center of the desired circle lies on the line x = 15. We also know, since said line is a bisector of the chord, that the distance from the point (6,0) to the point of intersection of the line x = 15 and the x-axis is 9. Since the circle is tangent to the y-axis, the distance from the y-axis to the center of the circle must be 15, hence the radius of the circle is 15.

Consider the right triangle formed by the radius of the circle that has the point (6,0) as an end point, the portion of the line x = 15 from the center of the circle to the center of the chord segment of the x-axis, and the half-chord from point (6,0) to (15,0). Since this is a right triangle with hypotenuse/short leg ratio of 5:3, the other leg must be in ratio 5:4:3 with the other two sides. Hence the distance from the x-axis to the center of the circle is 12.

Our circle is centered at (15,12) and has a radius of 15.

A circle centered at (h, k) with radius r has an equation:

John

My calculator said it, I believe it, that settles it

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