In the standard (x, y) coordinate plane, a circle with center (5, -1) is tangent to the x-… — ACT

In the standard $(x, y)$ coordinate plane, a circle with center $(5, -1)$ is tangent to the $x$ -axis at point $P$ . A line passes through point $P$ and the point $(-3, 6)$ . What is the slope of this line?

A
$-\frac{4}{3}$
B
$-\frac{7}{8}$
$-\frac{3}{4}$ Answer
D
$\frac{3}{4}$
E
$-\frac{7}{3}$

Answer

-\frac{3}{4}

The circle with center

(5, -1)

is tangent to the

x

-axis, meaning the radius is

1

and the point of tangency

P

lies on the

x

-axis directly above the center at

(5, 0)

. The line passes through

P(5, 0)

and

(-3, 6)

. Applying the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

gives

m = \frac{6 - 0}{-3 - 5} = \frac{6}{-8} = -\frac{3}{4}

Step-by-Step Solution

Determine the coordinates of the point of tangency

P

P = (5, 0)

Since the circle has center

(5, -1)

and is tangent to the

x

-axis, the point of tangency must lie on the

x

-axis (where

y = 0

). The point on the

x

-axis directly above the center

(5, -1)

(5, 0)

Set up the slope formula for the line passing through point

P(5, 0)

and point

(-3, 6)

m = \frac{6 - 0}{-3 - 5}

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by the formula

m = \frac{y_2 - y_1}{x_2 - x_1}

Simplify the expression to find the slope.

m = -\frac{3}{4}

Calculate the difference in the numerator and denominator:

\frac{6}{-8}

, then reduce the fraction to simplest form:

-\frac{3}{4}

Key Concept

Calculating the slope of a line using the coordinates of two points, where one point is determined by a geometric tangency property.

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