Write the standard equation of a circle that passes through (1, −6) with center (7, −2).

A: (x + 1)2 + (y – 6)2 = 52

B: (x + 7)2 + (y – 2)2 = 52

C: (x – 1)2 + (y + 6)2 = 52

D: (x – 7)2 + (y + 2)2 = 52

D: (x – 7)2 + (y + 2)2 = 52

Step-by-step explanation:

Circle: (x -h)² + (y - k)² = r²    center (h , k)    r: radius

(1 , -6) on the circle: radius   √(1 - 7)² + (-6 - (-2))² = √36 + 16 = √52

(x -7)² + (y +2)² = (√52)² = 52

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°

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