Chapter 9, Problem 2CT

To determine

To determine the equation of a circle with centre $(6,-3)$ that passes through point $(-4,1)$

Answer to Problem 2CT

  ${(x-6)}^2+{(y+3)}^2=116$ is the equation of the circle.

Explanation of Solution

Given:

A circle with centre $(6,-3)$ that passes through point $(-4,1)$

Formula Used:

The equation of a circle is:

  ${(x-a)}^2+{(y-b)}^2=r^2$ where $(a,b)$ is the centre of the circle and $r$ is the radius of the circle.

The formula for calculating the distance between the two points is:

  $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Calculation:

Given the centre $(6,-3)$ of the circle but we need to find the radius of the circle.

The radius is the distance between the centre of the circle and any point on the circle. Both the centre of the circle and the point on the circle is given; we have to calculate the distance between these two points which is the radius of the circle.

The formula for calculating the distance between the two points is:

  $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Substituting the values from the points gives:

  $r=\sqrt{{(-4-6)}^2+{(1-(-3))}^2}$

  $r=\sqrt{{(-10)}^2+{(4)}^2}$

  $r=\sqrt{116}$

Now, the equation of the circle is represented as:

  ${(x-a)}^2+{(y-b)}^2=r^2$

Substituting the values of a, b and r from above; we get:

  ${(x-6)}^2+{(y-(-3))}^2={(\sqrt{116})}^2$

Solving it further,

  ${(x-6)}^2+{(y+3)}^2=116$

Hence, the answer is ${(x-6)}^2+{(y+3)}^2=116$

Conclusion:

  ${(x-6)}^2+{(y+3)}^2=116$ is the solution.