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a circle in the xy - plane has its center at (2, 7). line t is tangent …

Question

a circle in the xy - plane has its center at (2, 7). line t is tangent to this circle at the point (a, - 4), where a is a constant. the slope of line t is -\frac{6}{5}. what is the value of a? a -\frac{56}{5} b -\frac{43}{6} c \frac{67}{6} d \frac{76}{5}

Explanation:

Step1: Find slope of radius

The slope of the line (radius) connecting the center of the circle $(2,7)$ and the point of tangency $(a, - 4)$ is given by the slope - formula $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-4 - 7}{a - 2}=\frac{-11}{a - 2}$.

Step2: Use property of tangent and radius

A tangent to a circle is perpendicular to the radius at the point of tangency. If two non - vertical lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1\times m_2=-1$. The slope of the tangent line $t$ is $\frac{6}{5}$. So, $\frac{-11}{a - 2}\times\frac{6}{5}=-1$.

Step3: Solve for $a$

Cross - multiply the equation $\frac{-11\times6}{5(a - 2)}=-1$ to get $-66=-5(a - 2)$.
Expand the right - hand side: $-66=-5a + 10$.
Add $5a$ to both sides: $5a-66 = 10$.
Add 66 to both sides: $5a=10 + 66=76$.
Divide both sides by 5: $a=\frac{76}{5}$.

Answer:

D. $\frac{76}{5}$