Question

a circle in the xy - plane has its center at (2, 9). 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 $-\frac{68}{5}$
b $-\frac{53}{6}$
c $\frac{77}{6}$
d $\frac{88}{5}$

Explanation:

Step1: Recall tangent - radius property

The radius of a circle is perpendicular to the tangent line at the point of tangency. If the slope of line \(t\) is \(m_t=\frac{6}{5}\), and the slope of the radius connecting the center \((2,9)\) and the point of tangency \((a, - 4)\) is \(m_r\), then \(m_t\times m_r=- 1\) (since perpendicular lines have slopes that are negative - reciprocals of each other). So, \(m_r=-\frac{5}{6}\).

Step2: Use the slope formula

The slope formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Here, \(x_1 = 2,y_1 = 9,x_2=a,y_2=-4\), and \(m=-\frac{5}{6}\). Substitute these values into the slope formula: \(-\frac{5}{6}=\frac{-4 - 9}{a - 2}\).

Step3: Simplify the equation

First, simplify the numerator on the right - hand side: \(-\frac{5}{6}=\frac{-13}{a - 2}\). Cross - multiply to get \(-5(a - 2)=-13\times6\).

Step4: Expand and solve for \(a\)

Expand the left - hand side: \(-5a+10=-78\). Subtract 10 from both sides: \(-5a=-78 - 10=-88\). Then divide both sides by \(-5\): \(a=\frac{88}{5}\).

Answer:

D. \(\frac{88}{5}\)