deduce the equation of the circle

deduce the equation of the circle

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Gabor R. | Experienced Math and Science TutorExperienced Math and Science Tutor

When a circle is tangent to a line, the line is perpendicular to the radius at the point of tangency. So since this circle is tangent to the y-axis at y=3, the y-coordinate of the center of the circle must also be 3. Let c_{x} be the x-coordinate of the center of this circle. c_{x} is also the radius, since the distance between the point of tangency (0,3) and the center of the circle (c_{x},3) is c_{x}.So by the Pythagorean Theorem we have

c_{x}^{2} = (c_{x} - 1)^{2 }+ (3 - 0)^{2}.

Hence 0 = -2c_{x} + 10. So c_{x} = 5. It follows that the equation of the circle is

(x - 5)^{2} + (y - 3)^{2} = 5^{2}.^{
}

Rather than using the equation of the circle to find the other x-intercept by brute force, let us use the fact that both the circle and the line it intersects (the x-axis) have mirror symmetry about the line x = c_{x}. So any feature (such as an intersection) of the two which appears at (x, y) must also appear at (c_{x}+ (c_{x }- x), y) = (10 - x, y). So since there is an x-intercept of x=1, i.e. appearing at (1, 0), there must be one at (9, 0), i.e. an x-intercept of x=9.

Ok, this one is a bit tricky to discuss without drawing a figure so bear with me.

The equation of a circle is (x -x_{0})^{2} + (y - y_{0})^{2} = r^{2} where (x_{0, }y_{0}) is the location of the circles center. We have three unknowns to find including the radius. Draw a bunch of circles tangent to the y-axis. Perpendicular to the point of tangency is a radius so that tells us that the height of the center is at the point of tangency so y_{0} = 3. So the equation is now (x -x_{0})^{2} + (y - 3)^{2} = r_{2}

Lets use another piece of information: at x = 0, y = 3 since the circle touches the y-axis.

(0 -x_{0})^{2} + (3 - 3)^{2} = r^{2 }or

x_{0}^{2} = r^{2}

So once we know the x-coordinate of the center, we'll know r.

Lets use the known x-intercept and plug the numbers into the equation for the circle.

(1 -x_{0})^{2} + (0 - 3)^{2} = r^{2}

(1 -x_{0})^{2} + 9 = r^{2}

In the equation we just derived, if we replace r^{2} with x_{0}^{2} using the last formula we derived, we'll get x0 = 5 and r = 5. So the equation is

(x - 5)^2 + (y - 3)^2 = 25

Plug the given point of tangency and the x-intercept into the formula as a check. Both sides should be equal to 25. Then, replace y with 0 and find the second x-intercept (x = 9)

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