Question

1.Find the circumcentre of the triangle wose vertices
are (-2,-3). (-1,0), (7-6). Also find the radius of circum circle.​

Answer 1

Answer :

Coordinates of circumcentre = (3, -3)

Step-by-step explanation :

Given that :

A triangle ABC, A(7,-6), B(-2,-3), C(-1,0) Circumcentre P ( x, y).

To Find :

The radius of circum circle.​

Solution :

We know that :

Section formula.

So,

$\tt \implies X_1 = \dfrac{(m1x2 + m2x1)}{2}, Y_1 = \dfrac{(m1y1 + m2y2)}{2}$

We have to use Distance formula too here, So,

$\tt \implies \sqrt{{( x1-x2)^2 + ( y1-y2)^2}$

$\tt \implies x1= \dfrac{-3}{2}\;and\;y1 = \dfrac{-3}{2}$

We also know :

$\bigstar$ PB² = BM² + PM²

Values on Equation,

$\tt \implies (x+2)^2+ (y+3)^2 =(-2+ \dfrac{3}{2})^2 + (-3 + \dfrac{3}{2})^2 + ( x+\dfrac{3}{2})^2 + (y + \dfrac{3}{2})^2\\ \\\implies x^2 + 4 + 4x+ y^2 +9 + 6y= \dfrac{1}{4}1+ \dfrac{9}{4}+ +x^2 + \dfrac{9}{4} + 3x y^2 + \dfrac{9}{4} + 3y\\ \\\implies 4x + 6y + 13 = 7 + 3x + 3y\\ \\\implies x + 3y = -6 .....Eq(1)$

$\sf{The\; midpoint\;AB\; is\; N, the\; coordinates\;of \;which\; x_2 = \dfrac{(-2+7)}{2} = \dfrac{5}{2}}\\ \\\implies y_2 = \dfrac{(-3 + -6)}{2} = \dfrac{-9}{2}$

We know that :

Pythagoras law in triangle PAN

PA² = AN² = + PN²

Plug the values :

$\tt \implies (x-7)^2 + (y+6)^2 = (7- \dfrac{5}{2})^2 + (-6+\dfrac{9}{2})^2 + (x-\dfrac{5}{2})^2 + (y+\dfrac{9}{2})^2\\ \\\implies x^2 + 49 - 14x +y^2 +36 + 12y = \dfrac{81}{4} + \dfrac{9}{4} + x^2 + \dfrac{25}{4} - 5x + y^2 + 9y + \dfrac{81}{4}\\ \\\implies 85 -14x + 12y = 49 -5x + 9y\\ \\\implies -9x + 3y = -36\\ \\\implies -3x + y = -12 ......Eq(2)$

$\textbf{\underline{\underline{Now, Solving Eq(1) and Eq(2)}}}\\ \\ \\\sf \implies x+ 3y =-6....(1)\\ \\\implies -3x + y = -12.....(2)\\ \\\implies x = 3\\ \\\implies y = -3\\ \\ \\\bigstar\;\textbf{\underline{\underline{So, coordinates of circumcentre = (3, -3)}}}$

Answer 2

very easy solution.

Thanks .