Question

1. find the center of the circle that can be circumscribed about △abc.

(6, 4)
(4, 0)
(3, 2)
(0, 0)

Explanation:

Step1: Recall circum - center property

The circum - center of a triangle is equidistant from all the vertices of the triangle. Let \(A=(0,0)\), \(B=(0,4)\) and \(C=(6,4)\).

Step2: Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Let the circum - center be \(O=(x,y)\). Then \(OA=\sqrt{(x - 0)^2+(y - 0)^2}=\sqrt{x^{2}+y^{2}}\), \(OB=\sqrt{(x - 0)^2+(y - 4)^2}=\sqrt{x^{2}+(y - 4)^{2}}\) and \(OC=\sqrt{(x - 6)^2+(y - 4)^2}\).
Since \(OA = OB\), we have \(\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+(y - 4)^{2}}\). Squaring both sides gives \(x^{2}+y^{2}=x^{2}+y^{2}-8y + 16\), which simplifies to \(8y=16\) and \(y = 2\).
Since \(OB=OC\), substituting \(y = 2\) into \(OB\) and \(OC\) expressions: \(OB=\sqrt{x^{2}+(2 - 4)^{2}}=\sqrt{x^{2}+4}\), \(OC=\sqrt{(x - 6)^2+(2 - 4)^2}=\sqrt{(x - 6)^{2}+4}\).
Setting \(\sqrt{x^{2}+4}=\sqrt{(x - 6)^{2}+4}\), squaring both sides: \(x^{2}+4=(x - 6)^{2}+4\). Expanding \((x - 6)^{2}=x^{2}-12x + 36\), so \(x^{2}+4=x^{2}-12x + 36+4\). Canceling out \(x^{2}\) and \(4\) on both sides gives \(12x=36\), and \(x = 3\).

Answer:

C. \((3,2)\)