Identify an equation in point-slope form for the line parallel to y = - - x
EX+8 that
passes through (4, -5).

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Step-by-step explanation:

the answer is (2x+8)²= 16

< 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

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