# How do you find the center of the circle that can be circumscribed about ΔEFG with E (2,2), F (2,-2), and G (6,-2)?

Sep 14, 2016

$\text{ The circumcentre is } \left(4 , 0\right)$.

#### Explanation:

Observe that the pts. $E \left(2 , 2\right) \mathmr{and} F \left(2 , - 2\right)$ have the same x-co-

ordinates.

$\text{:. line "EF" is vertical", i.e., ||" to "Y"-axis}$.

Similarly, $\text{line "FG" is horizontal, i.e.,"||" to "X"-axis}$.

Therefore, m/_EFG=90^@, &, "in "DeltaEFG, EG" is hypotenuse."

We know from Geometry that, in a right-$\Delta$, the mid-pt. of the

hypotenuse is the circumcentre of the $\Delta$.

$\therefore \text{ The circumcentre is the mid-pt. of seg. } E G = \left(\frac{2 + 6}{2} , \frac{2 - 2}{2}\right)$

$= \left(4 , 0\right)$.

Enjoy Maths.!