Find the equation of a circle with a center at (0,-4) and a point on the circle is (6,0). O( x-6 )2 ( + y ) = 52
O( x )2 +( y +4 )' = 52
O( x-6)2 + y )' = 152
O( x )2 +( y +4 ) = 152

< 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⭐

c 1 hour

step-by-step explanation:

m = 2e

m = 2 so 2 = 2e

1 = e

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3