Question

An equation of circle
H
H is shown.



(
x
+
4
)
2
+
(
y
+
4
)
2
=
100
(x+4)
2
+(y+4)
2
=100



Line
K
L
KL passes through the points
K
(
18
,
−
8
)
K(18,−8) and
L
(
2
,
4
)
L(2,4) and is tangent to the circle at
(
2
,
4
)
,
(2,4), as shown.







What is the equation of the line that is parallel to
K
L
↔
KL
and is also tangent to circle
H
H at point
(
−
10
,
−
12
)
(−10,−12)?

Answer 1

To find the equation of the line that is parallel to line \( KL \) and tangent to circle \( H \) at the point \((-10, -12)\), we first need to determine the slope of line \( KL \) and then use that slope to find the equation of the tangent line.

### Step 1: Find the slope of line \( KL \)

We have points \( K(18, -8) \) and \( L(2, 4) \). The slope \( m \) of line \( KL \) can be calculated using the formula for the slope between two points:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substituting in the coordinates of points \( K \) and \( L \):

\[
m = \frac{4 - (-8)}{2 - 18} = \frac{4 + 8}{2 - 18} = \frac{12}{-16} = -\frac{3}{4}
\]

### Step 2: Find the slope of the tangent line

Since we want a line parallel to \( KL \), it will have the same slope:

\[
m = -\frac{3}{4}
\]

### Step 3: Use the point-slope form to find the equation of the tangent line

We will use the point \((-10, -12)\) where the tangent line touches the circle. The point-slope form of a linear equation is:

\[
y - y_1 = m(x - x_1)
\]

Substituting in the point \((-10, -12)\) and the slope \(-\frac{3}{4}\):

\[
y - (-12) = -\frac{3}{4}(x - (-10))
\]
\[
y + 12 = -\frac{3}{4}(x + 10)
\]

### Step 4: Simplify the equation

We can distribute and rearrange the equation:

\[
y + 12 = -\frac{3}{4}x - \frac{3}{4} \cdot 10
\]
\[
y + 12 = -\frac{3}{4}x - \frac{30}{4}
\]
\[
y + 12 = -\frac{3}{4}x - \frac{15}{2}
\]

Now, we isolate \( y \):

\[
y = -\frac{3}{4}x - \frac{15}{2} - 12
\]
\[
y = -\frac{3}{4}x - \frac{15}{2} - \frac{24}{2}
\]
\[
y = -\frac{3}{4}x - \frac{39}{2}
\]

### Final Answer

The equation of the line that is parallel to \( KL \) and tangent to circle \( H \) at the point \((-10, -12)\) is:

\[
y = -\frac{3}{4}x - \frac{39}{2}
\]