Given that the endpoints of a diameter are (7,-3) and (-1,7), determine the equation of the circle.

The equation of a circle is written as (x-h)^2 + (y-k)^2 = r^r

H and k are the x and y coordinates of the center of the circle and r is the radius.

You are given the diameter coordinates so find the halfway point for the center then calculate the radius

Midpoint = (x1 +x2)/2, (y1 + y2)/2

Midpoint = (7 + -1)/2, (-3 +7)/2

Midpoint = 6/2, 4/2

Midpoint = 3,2

So h = 3 and k = 2

Now find radius by finding the distance between the center point and an end point.

Distance = sqrt(41)

Equation of the circle:

(X-3)^2 + (y-2)^2 = 41

The equation of the circle is written as:

(x-h)²+(y-k)²=r²

'h' and 'k' are the x and y coordinates of the center of the circle and 'r' is the radius.

You are given that the diameter coordinates, so find the halfway point for the center then calculate the radius

So, h=3 and k=2

Now find the radius by finding the distance between the center point and the end point.

Distance=√{(7-3)²+(-3-2)²}=√{(4)²+(-5)²} =√{16+25}=√41

Equation of the circle:

(x-3)²+(y-2)²=41 ✓

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.

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )