Finding Equation - Circle, 8

Category: Analytic Geometry, Plane Geometry, Algebra

"Published in Newark, California, USA"

Find the equation of a circle which is tangent to the y-axis and passes through the points (1, 5) and (8, -2).

Solution:

To illustrate the problem, it is better to draw the figure as follows

A circle is tangent to the y-axis and passes thru (1, 5) and (8, -2). (Photo by Math Principles in Everyday Life)

Since the given circle is tangent to the y-axis, then the radius is equal to h which is the x-coordinate of the center of a circle. 

Since the two points of a circle are given, then we can use the distance of two points formula in order to get the radius of a circle as follows

We need to get another equation because the above equation consists of two variables as follows

Substitute the value of h from equation (1) to equation (2), we have

After equating each factor to zero, the values of k are 10 and 2.

If k = 10, then

Hence, the center of a circle is C (13, 10) and its radius is r = h = 13.

Therefore, the equation of a circle is

If k = 2, then

Hence, the center of a circle is C (5, 2) and its radius is r = h = 5.

Therefore, the equation of a circle is

Finding Equation - Circle, 7

Category: Analytic Geometry, Plane Geometry, Algebra

"Published in Newark, California, USA"

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

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

and

 

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