Write the equation of the circle with Center (-1, -8) and Radius: 7 a) x2+2x+y2+16y+58=0 b) x2+2x+1+y2+16y+64=7 c) x2+2x+y2+16y+72=0 d) x2−2x+y2−16y+58=0

Step 1 Given point on circle ⇒(a, b)=(−1, −8) and radius ⇒r=7 As we know given equation of circle is i.e (x−a)2+(y−b}2=r2 or (x+1)2+(y+8)2=(7)2=7 x2+1+2x+y2+64+16y=7 x2+y2+2x+16y+58=0 Hence option (a) is correct

Question: "Write the equation of the circle with Center..."

berljivx8

Answered question

2021-12-13

Write the equation of the circle with Center (-1, -8) and Radius:

\sqrt{7}

x^{2} + 2 x + y^{2} + 16 y + 58 = 0

x^{2} + 2 x + 1 + y^{2} + 16 y + 64 = 7

x^{2} + 2 x + y^{2} + 16 y + 72 = 0

x^{2} - 2 x + y^{2} - 16 y + 58 = 0

Answer & Explanation

Mary Nicholson

Beginner2021-12-14Added 38 answers

Step 1
Given point on circle

\Rightarrow (a, b) = (- 1, - 8)

and radius

\Rightarrow r = \sqrt{7}

As we know given equation of circle is i.e

{(x - a)}^{2} + {(y - b}}^{2} = r^{2}

{(x + 1)}^{2} + {(y + 8)}^{2} = {(\sqrt{7})}^{2} = 7

x^{2} + 1 + 2 x + y^{2} + 64 + 16 y = 7

x^{2} + y^{2} + 2 x + 16 y + 58 = 0

Hence option (a) is correct

Samantha Brown

Beginner2021-12-15Added 35 answers

Step 1
Recall that the equation of a circle is defined as:

{(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}

Where

(x_{0}, y_{0})

is the center of the circle.
Given the data we have, we know that the radius of the circler must be 2(half the diameter). Thus, we know question must be;

{(x - (- 1))}^{2} + {(y - (- 8))}^{2} = \sqrt{7}

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