MATH SOLVE

4 months ago

Q:
# 4. Find the center and the radius of the circle which circumscribes the triangle with vertices ai, a, a3. Express the result in symmetric form.

Accepted Solution

A:

Answer:[tex]\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]=\left[\begin{array}{ccc}-a_{1}^{2}-b_{1}^{2}\\-a_{2}^{2}-b_{2}^{2}\\-a_{3}^{2}-b_{3}^{2}\end{array}\right][/tex]Step-by-step explanation:In the question,We have to find out the circumcentre of the circle passing through the triangle with the vertices (aβ, bβ), (aβ, bβ) and (aβ, cβ).So,The circle is passing through these points the equation of the circle is given by,[tex]x^{2}+y^{2}+ax+by+c=0[/tex]On putting the points in the circle we get,[tex]x^{2}+y^{2}+ax+by+c=0\\(a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\and,\\(a_{2})^{2}+(b_{2})^{2}+a(a_{2})+b(b_{2})+c=0\\and,\\(a_{3})^{3}+(b_{3})^{3}+a(a_{3})+b(b_{3})+c=0\\[/tex]So,[tex](a_{1})^{2}+(b_{1})^{2}+a(a_{1})+b(b_{1})+c=0\\a(a_{1})+b(b_{1})+c=-(a_{1})^{2}-(b_{1})^{2}\,.........(1)\\and,\\a(a_{2})+b(b_{2})+c=-(a_{2})^{2}-(b_{2})^{2}\,.........(2)\\and,\\a(a_{3})+b(b_{3})+c=-(a_{3})^{3}-(b_{3})^{3}\,.........(3)\\[/tex]On solving these equation using, Matrix method we can get the required equation of the circle,[tex]\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]=\left[\begin{array}{ccc}-a_{1}^{2}-b_{1}^{2}\\-a_{2}^{2}-b_{2}^{2}\\-a_{3}^{2}-b_{3}^{2}\end{array}\right][/tex]This is the required answer.