Question

Find the equation of a circle with its center at (3, 7) and tangent to the line 3x + 2y - 5 = 0

Answer 1

To find the equation of a circle with a given center and tangent to a given line, we can follow these steps:

1. **Identify the center and the equation of the line**: The center of the circle is given as \( (3, 7) \), and the equation of the line is \( 3x + 2y - 5 = 0 \).

2. **Find the distance from the center of the circle to the line**: The distance \( d \) from a point \( (x_0, y_0) \) to a line given by \( Ax + By + C = 0 \) can be calculated using the formula:
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
In our case, \( A = 3 \), \( B = 2 \), and \( C = -5 \) (from the equation \( 3x + 2y - 5 = 0 \)). The point is \( (3, 7) \).

3. **Substitute the values into the distance formula**:
\[
d = \frac{|3(3) + 2(7) - 5|}{\sqrt{3^2 + 2^2}} = \frac{|9 + 14 - 5|}{\sqrt{9 + 4}} = \frac{|18|}{\sqrt{13}} = \frac{18}{\sqrt{13}}
\]

4. **The radius of the circle**: Since the circle is tangent to the line, the distance \( d \) we calculated is the radius \( r \) of the circle:
\[
r = \frac{18}{\sqrt{13}}
\]

5. **Formulate the equation of the circle**: The standard equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In this case, \( h = 3 \), \( k = 7 \), and \( r^2 = \left(\frac{18}{\sqrt{13}}\right)^2 = \frac{324}{13} \).

6. **Write the final equation of the circle**:
\[
(x - 3)^2 + (y - 7)^2 = \frac{324}{13}
\]

Thus, the equation of the circle is:
\[
(x - 3)^2 + (y - 7)^2 = \frac{324}{13}
\]