To find the equation of the tangent line to the circle at point [tex]\( Q \)[/tex], we need to determine the slope of the line tangent to the circle at [tex]\( Q \)[/tex]. Here are the steps to solve the problem:
1. Identify the slope of [tex]\(\overline{PQ}\)[/tex]:
Given the line equation [tex]\( y = \frac{2}{5}x + 3 \)[/tex], we see that the slope [tex]\( m \)[/tex] of this line is [tex]\( \frac{2}{5} \)[/tex].
2. Determine the slope of the tangent line:
The tangent line at point [tex]\( Q \)[/tex] on the circle will be perpendicular to the radius [tex]\(\overline{PQ}\)[/tex]. The slope of a line perpendicular to another line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex].
Therefore, the slope of the tangent line [tex]\( m_t \)[/tex] will be:
[tex]\[
m_t = -\frac{1}{m} = -\frac{1}{\frac{2}{5}} = -\frac{5}{2}
\][/tex]
3. Identify which given equations have the slope [tex]\(-\frac{5}{2}\)[/tex]:
We need to write each given linear equation in the slope-intercept form [tex]\( y = mx + b \)[/tex] to compare slopes.
- Equation 1: [tex]\( 2x + 5y = 1 \)[/tex]
[tex]\[
5y = -2x + 1 \implies y = -\frac{2}{5}x + \frac{1}{5}
\][/tex]
The slope is [tex]\(-\frac{2}{5}\)[/tex].
- Equation 2: [tex]\( 2x + y = 3 \)[/tex]
[tex]\[
y = -2x + 3
\][/tex]
The slope is [tex]\(-2\)[/tex].
- Equation 3: [tex]\( 5x + 2y = 7 \)[/tex]
[tex]\[
2y = -5x + 7 \implies y = -\frac{5}{2}x + \frac{7}{2}
\][/tex]
The slope is [tex]\(-\frac{5}{2}\)[/tex].
- Equation 4: [tex]\( 2x - 5y = 2 \)[/tex]
[tex]\[
-5y = -2x + 2 \implies y = \frac{2}{5}x - \frac{2}{5}
\][/tex]
The slope is [tex]\(\frac{2}{5}\)[/tex].
4. Select the correct equation:
The slope of the tangent line should be [tex]\(-\frac{5}{2}\)[/tex]. The only equation with this slope is [tex]\( 5x + 2y = 7 \)[/tex] from equation 3.
Therefore, the correct answer is:
[tex]\[ 5x + 2y = 7 \][/tex]
