What is the radius of a circle represented by the equation x²+y²x²+y²-2x+4y-18=0

Step-by-step explanation:

i think eq. of circle is

x²+y²-2x+4y-18=0

compare with x²+y²+2gx+2fy+c=0

center is (-g,-f) or (1,-2)

radius is √(g²+f²-c)=√((1)²+(-2)²-(-18))

=√(1+4+18)

=√23

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!