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 (a₁, b₁), (aβ‚‚, bβ‚‚) and (a₃, c₃).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.