(x-h)^2+(y-k)^2=r^2 Derive standard form of an equation of a circle with center C(5,2) and radius r=4, using the Pythagorean theorem.

(x-2)^2 + (y - 5)^2 = 16

Step-by-step explanation:

Given the standard form of an equation of a circle

(x-h)^2+(y-k)^2=r^2

If (h, k) = (5,2) and r = 4

Substitute

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

(x-2)^2 + (y - 5)^2 = 16

Hence the required equation is (x-2)^2 + (y - 5)^2 = 16

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°