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Find the equation of a circle that passes through the point (2, 2), and tangent to the lines x = 1 and x = 6.
Solution:
To illustrate the problem, it is better to draw the figure as follows
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| There are two possible circles that passes thru the point (2,2). (Photo by Math Principles in Everyday Life) |
As you can see from the figure that there are two possible equations of a circle that is tangent to x = 1 and x = 6 and passes through the point (2, 2).
To solve for the coordinates of the center of a circle, we need to use the perpendicular distance of a point to the line formula
and the distance of two points formula
In this problem, d will be equal to r which is the radius of a circle and we need to get the value of radius also.
If a circle is tangent to x = 1, then
Substitute the values of A, B, and C which is the coefficient of x = 1 or x - 1 = 0, we have
Since the sign of 1 is negative, then the sign of a radical is positive. We need to take the opposite sign of the coefficient.
Substitute the values of the coordinates of the center of a circle which is C (h, k) to the right side of the equation, we have
Substitute the values of the endpoints of the radius of a circle, we have
Since x = 1 and x = 6 are parallel to each other and they are perpendicular to x-axis, then the center of a circle is located along x = 3.5 or x = 7/2 which is the median of two parallel lines. If a circle is tangent to x = 1 and x = 6, then the coordinates of a center is C (7/2, k). Substitute the value of h to the above equation in order to solve for the value of k, we have
Equate each factor and solve for the value of k, we have
Hence, the center of a circle is C (7/2, 0) or C (7/2, 4).
Since a circle is tangent to x = 1 and x = 6, then it follows that the diameter of a circle is 5 which is the distance of two parallel line. The radius of a circle is 2.5 or 5/2.
Therefore, the equation of a circle with C (7/2, 0) is
If the center of a circle is C (7/2, 4), then the equation of a circle is
