1. **State the problem:** Find the equation of the inscribed circle (incircle) of the triangle with vertices $A(-10,6)$, $B(10,8)$, and $C(0,-8)$. 2. **Recall the formula for a circle:** The equation of a circle with center $(h,k)$ and radius $r$ is $$ (x - h)^2 + (y - k)^2 = r^2 $$ 3. **Identify the center:** The incircle center (incenter) is given approximately as $I(0,2)$. 4. **Find the radius $r$:** The radius is the distance from the incenter to any side of the triangle (the perpendicular distance). 5. **Calculate the distance from $I$ to side $AB$:** - Equation of line $AB$: Slope $m = \frac{8 - 6}{10 - (-10)} = \frac{2}{20} = \frac{1}{10}$ Using point-slope form with point $A(-10,6)$: $$ y - 6 = \frac{1}{10}(x + 10) \implies y = \frac{1}{10}x + 7 $$ Rewrite in standard form: $$ y - \frac{1}{10}x - 7 = 0 \implies -\frac{1}{10}x + y - 7 = 0 $$ Multiply by 10 to clear fractions: $$ -x + 10y - 70 = 0 $$ 6. **Distance from point $I(0,2)$ to line $AB$:** $$ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|-1 \cdot 0 + 10 \cdot 2 - 70|}{\sqrt{(-1)^2 + 10^2}} = \frac{|20 - 70|}{\sqrt{1 + 100}} = \frac{50}{\sqrt{101}} $$ 7. **Calculate the distance from $I$ to side $BC$:** - Equation of line $BC$: Slope $m = \frac{8 - (-8)}{10 - 0} = \frac{16}{10} = \frac{8}{5}$ Using point $B(10,8)$: $$ y - 8 = \frac{8}{5}(x - 10) \implies y = \frac{8}{5}x - 16 + 8 = \frac{8}{5}x - 8 $$ Rewrite in standard form: $$ y - \frac{8}{5}x + 8 = 0 \implies -\frac{8}{5}x + y + 8 = 0 $$ Multiply by 5: $$ -8x + 5y + 40 = 0 $$ Distance from $I(0,2)$: $$ d = \frac{|-8 \cdot 0 + 5 \cdot 2 + 40|}{\sqrt{(-8)^2 + 5^2}} = \frac{|10 + 40|}{\sqrt{64 + 25}} = \frac{50}{\sqrt{89}} $$ 8. **Calculate the distance from $I$ to side $CA$:** - Equation of line $CA$: Slope $m = \frac{6 - (-8)}{-10 - 0} = \frac{14}{-10} = -\frac{7}{5}$ Using point $C(0,-8)$: $$ y + 8 = -\frac{7}{5}(x - 0) \implies y = -\frac{7}{5}x - 8 $$ Rewrite in standard form: $$ y + \frac{7}{5}x + 8 = 0 \implies \frac{7}{5}x + y + 8 = 0 $$ Multiply by 5: $$ 7x + 5y + 40 = 0 $$ Distance from $I(0,2)$: $$ d = \frac{|7 \cdot 0 + 5 \cdot 2 + 40|}{\sqrt{7^2 + 5^2}} = \frac{|10 + 40|}{\sqrt{49 + 25}} = \frac{50}{\sqrt{74}} $$ 9. **Determine the radius:** The radius is the minimum of these distances because the incircle touches all sides. $$ r = \min\left(\frac{50}{\sqrt{101}}, \frac{50}{\sqrt{89}}, \frac{50}{\sqrt{74}}\right) = \frac{50}{\sqrt{101}} \approx 4.975 $$ 10. **Write the equation of the incircle:** $$ (x - 0)^2 + (y - 2)^2 = \left(\frac{50}{\sqrt{101}}\right)^2 = \frac{2500}{101} $$ **Final answer:** $$ \boxed{ x^2 + (y - 2)^2 = \frac{2500}{101} } $$