Question

Circle
T
T is shown with radius of
2
5
.

N
C
↔
2
5
​
.
NC
is tangent to the circle at point
H
(
−
1
,
0
)
.
H(−1,0). The equation of
N
C
↔
NC
is
2
x
+
y
=
−
2.
2x+y=−2.







Complete the equation of
F
T
↔
.
FT
.

Answer 1

To find the equation of the line \( FT \), which is the radius of the circle that extends from the center \( T \) to the point of tangency \( H \), we first need to identify several key pieces of information:

1. **Identify the center of the circle \( T \)**: Since the radius of the circle is \( 2.5 \) (or \( 5/2 \)), we can denote \( T \) as (h, k). We know that point \( H \) is the point of tangency (−1, 0) and that \( H \) lies on the circle.

2. **Using the distance formula to find the center \( T \)**:
\[
\sqrt{(h - (-1))^2 + (k - 0)^2} = 2.5.
\]
Therefore:
\[
(h + 1)^2 + k^2 = (2.5)^2 = 6.25.
\]

3. **Find the slope of line \( NC \)**:
The equation of line \( NC \) is \( 2x + y = -2 \). We can rearrange it to the slope-intercept form:
\[
y = -2x - 2,
\]
which implies the slope of \( NC \) is \( -2 \).

4. **Determine the slope of the radius \( FT \)**:
Since \( FT \) is perpendicular to the tangent line \( NC \), the slope of \( FT \) will be the negative reciprocal of the slope of \( NC \):
\[
\text{slope of } FT = \frac{1}{2}.
\]

5. **Use the slope and the point \( H \)** to find the equation of line \( FT \)**:
We can use the point-slope form of the line equation:
\[
y - y_1 = m(x - x_1),
\]
where \( m = \frac{1}{2} \) and \( (x_1, y_1) = (-1, 0) \):
\[
y - 0 = \frac{1}{2}(x - (-1)),
\]
or
\[
y = \frac{1}{2}(x + 1).
\]
Simplifying this gives:
\[
y = \frac{1}{2}x + \frac{1}{2}.
\]

Thus, the complete equation of the radius \( FT \) is:
\[
y = \frac{1}{2}x + \frac{1}{2}.
\]

If needed, you can also convert it to standard form:
\[
2y = x + 1 \quad \text{or} \quad x - 2y + 1 = 0.
\]