Question
5. Find the equation of the circle inscribed in the triangle with vertices \((-7,-10),(-7,15)\) and \((5,-1)\).
Ask by Chambers Weston.
Jan 27,2025 13:50
UpStudy AI Solution
Tutor-Verified Answer
Answer
The equation of the inscribed circle is \( (x + 2)^2 + y^2 = 25 \).
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To find the equation of the circle inscribed in the triangle with vertices \((-7,-10)\), \((-7,15)\), and \((5,-1)\), we first calculate the lengths of the sides of the triangle. The vertices can be labeled as \(A = (-7,-10)\), \(B = (-7,15)\), and \(C = (5,-1)\).
1. Calculate the lengths of the sides:
- \(AB = \sqrt{((-7) - (-7))^2 + (15 - (-10))^2} = \sqrt{0 + 25} = 5\)
- \(BC = \sqrt{(5 - (-7))^2 + ((-1) - 15)^2} = \sqrt{(12)^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20\)
- \(CA = \sqrt{((-7) - 5)^2 + ((-10) - (-1))^2} = \sqrt{(-12)^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15\)
2. Using the lengths of the sides, we find the semi-perimeter \(s\):
\[
s = \frac{AB + BC + CA}{2} = \frac{5 + 20 + 15}{2} = 20
\]
3. Now, calculate the coordinates of the incenter \((I_x, I_y)\) using the formula:
\[
I_x = \frac{aA_x + bB_x + cC_x}{a + b + c}, \quad I_y = \frac{aA_y + bB_y + cC_y}{a + b + c}
\]
Where \(a = BC\), \(b = CA\), and \(c = AB\).
Substituting the values:
- \(A_x = -7\), \(A_y = -10\)
- \(B_x = -7\), \(B_y = 15\)
- \(C_x = 5\), \(C_y = -1\)
- \(a = 20\), \(b = 15\), \(c = 5\)
We calculate:
\[
I_x = \frac{20(-7) + 15(-7) + 5(5)}{20 + 15 + 5} = \frac{-140 - 105 + 25}{40} = \frac{-220}{40} = -5.5
\]
\[
I_y = \frac{20(-10) + 15(15) + 5(-1)}{20 + 15 + 5} = \frac{-200 + 225 - 5}{40} = \frac{20}{40} = 0.5
\]
4. The inradius \(r\) can be computed using the formula:
\[
r = \frac{A}{s}
\]
where \(A\) is the area of the triangle. The area can be found using the determinant formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting:
\[
A = \frac{1}{2} \left| (-7)(15 - (-1)) + (-7)((-1) - (-10)) + 5((-10) - 15) \right|
\]
\[
= \frac{1}{2} \left| (-7)(16) + (-7)(9) + 5(-25) \right| = \frac{1}{2} \left| -112 - 63 - 125 \right| = \frac{1}{2} \left| -300 \right| = 150
\]
Thus,
\[
r = \frac{150}{20} = 7.5
\]
5. The equation of the circle is given by:
\[
(x - I_x)^2 + (y - I_y)^2 = r^2
\]
Plugging in the values:
\[
\left(x + 5.5\right)^2 + \left(y - 0.5\right)^2 = (7.5)^2
\]
\[
\left
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