Question

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.

Answer 1

The center of the circumscribed circle of the symmetric isosceles triangle is at the origin, and the radius is equal to the length of the triangle's equal sides, denoted as r.

Step-by-step explanation:

To find the center and the radius of the circle which circumscribes the triangle with vertices at ai, a, and a3, we must first understand the nature of the triangle. Given that the triangle is described to be symmetric with equal sides AB = BC = r, it is an isosceles triangle. The perpendicular bisector of the base a will pass through the midpoint of the base and the opposite vertex, given it is symmetric about this bisector. This will also be the diameter of the circumscribed circle. Consequently, as the triangle is isosceles and symmetric, we can use the properties of similar isosceles triangles to solve for the center and radius of the circumscribed circle.

Since the base a will also be the diameter of the circumscribed circle, and we know from geometry that the diameter is twice the radius (a = 2r), the radius of the circumscribing circle is r. The center of this circle is at the midpoint of the base a in the given symmetric form. Therefore, the center of the circle is at the origin due to the symmetric property and the radius remains r.

Answer 2

`\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]`

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), (a₂, b₂) and (a₃, c₃).

So,

The circle is passing through these points the equation of the circle is given by,

`x^(2)+y^(2)+ax+by+c=0`

On putting the points in the circle we get,

`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\\`

So,

`(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)\\`

On solving these equation using, Matrix method we can get the required equation of the circle,

`\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]`

This is the required answer.