Mathematics, 25.09.2020 14:01, bren27

Please help me to solve the question!

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answered: malik1885

(x - 2)² + (y - 3)² = 25

Step-by-step explanation:

The equation of a circle is: (x - h)² + (y - k)² = r² where

(h, k) is the center of the circle"r" is the radius of the circle

We are given three coordinates. Input those as (x, y) into the circle equation. This creates a system of equations with 3 equations and 3 unknowns (h, k, r).

EQ1: (5 - h)² + (7 - k)² = r²

EQ2: (-1 - h)² + (7 - k)² = r²

EQ3: (5 - h)² + (-1 - k)² = r²

Evaluate EQ1 & EQ2:

(5 - h)² + (7 - k)² = r²

-[ (-1 - h)² + (7 - k)² = r² ]

(5 - h)² - (-1 - h)² = 0

(5 - h)² = (-1 - h)²

$\sqrt{(5 - h)^2} = \sqrt{(-1 - h)^2}$

5 - h = ±(-1 - h)

5 - h = -1 - h 5 - h = -(-1 - h)

6 ≠ 0 5 - h = 1 + h

Disregard 4 = 2h

2 = h

Input h = 2 into EQ2 & EQ3 and evaluate:

(-1 - 2)² + (7 - k)² = r² --> 9 + (7 - k)² = r²

(5 - 2)² + (-1 - k)² = r² --> - [ 9 + (-1 - k)² = r² ]

(7 - k)² - (-1 - k)² = 0

(7 - k)² = (-1 - k)²

$\sqrt{(7 - k)^2} = \sqrt{(-1 - k)^2}$

7 - k = ±(-1 - k)

7 - k = -1 - k 7 - k = -(-1 - k)

8 ≠ 0 7 - k = 1 + k

Disregard 6 = 2k

3 = k

Input h = 2 and k = 3 into any of the equations to solve for r²:

EQ1: (5 - 2)² + (7 - 3)² = r²

9 + 16 = r²

25 = r²

Input h = 2, k = 3, and r² = 25 into the equation of a circle:

(x - h)² + (y - k)² = r²

(x - 2)² + (y - 3)² = 25

answered: Guest

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step-by-step explanation:

answered: Guest

step-by-step explanation:

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