Which equation describes the circle having center point (3,5) and radius r = 2
in standard form?

x^2 + y^2 - 6x  - 10y + 30 = 0.

Step-by-step explanation:

In factor form the general equation is:

(x - h)^2 + (y - k)^2 = r^2 where  (h, k) is the centre and r = the radius.

So here we have:

(x - 3)^2 + (y - 5)^2 = 4.

Converting to Standard Form:

= x^2 - 6x + 9 + y^2 - 10y + 25 = 4

x^2 + y^2 - 6x  - 10y + 9 + 25 - 4 = 0

x^2 + y^2 - 6x  - 10y + 30 = 0.

< 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⭐

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes