Find the center of the circle that can be circumscribed about triangle efg with e(2, 2), f(2, -2), and g(6,-2).

(2, -1)

(4, 0)

(0, 4)

(3, 2)

The centre of circle is (4,0)          

Step-by-step explanation:

Given points are,

E(2, 2), F(2, -2), and G(6,-2).

The center of the circle is the circumcenter of the triangle, the point where the perpendicular bisectors intersect.

To find out the circumcenter we have to solve any two equations of bisectors of sides triangle.

Slope of the bisector is negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of EF with slope 0 and the coordinates (2,0) is,

  →    (1)

Similarly, for EG

So, the slope of the perpendicular bisector = 1

Equation of EG with slope 1 and the coordinates (4,0) is,

  →    (2)

Solving equation (1) and (2),

So the circumcenter is (4,0)

Option 2 is correct.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

x + 4/7

step-by-step explanation:

use cymath