Answer:
1) x² + y² - 4x + 12y - 20 = 0 ⇒ (x - 2)² + (y + 6)² = 60
2) x² + y² + 6x - 8y - 10 = 0 ⇒ No choice
3) 3x² + 3y² + 12x + 18y - 15 = 0 ⇒ (x + 2)² + (y + 3)² = 18
4) 5x² + 5y² - 10x + 20y - 30 = 0 ⇒ No choice
5) 2x² + 2y² - 24x - 16y - 8 = 0 ⇒ (x - 6)² + (y - 4)² = 56
6) x² + y² + 2x - 6y - 9 = 0 ⇒ (x + 1)² + (y - 6)² = 46
Step-by-step explanation:
- The general form of the equation of the circle is:
* x² + y² + Dx + Ey + F = 0
where D , E and F are constant
- The standard form of the equation of the circle is:
* (x - h)² + (y - k)² = r²
where (h , k) is the center of the circle, r is the radius of it
- To chose the circle equations in general form with their
corresponding equations in standard form lets do that
1) x² + y² - 4x + 12y - 20 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(-4)/2(1) = 2
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(12)/2(1) = -6
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (2)² + (-6)² - (-20) = 4 + 36 + 20 = 60
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x - 2)² + (y + 6)² = 60 ⇒ x² + y² - 4x + 12y - 20 = 0
2) x² + y² + 6x - 8y - 10 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(6)/2(1) = -3
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(-8)/2(1) = 4
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (-3)² + (4)² - (-10) = 9 + 16 + 10 = 35
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x + 3)² + (y - 4)² = 35 ⇒ there is no choice
3) 3x² + 3y² + 12x + 18y - 15 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(12)/2(3) = -2
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(18)/2(3) = -3
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (-2)² + (-3)² - (-15/3) = 4 + 9 + 5 = 18
- We divide F by 3 because the coefficient of x² and y²
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x + 2)² + (y + 3)² = 18 ⇒ 3x² + 3y² + 12x + 18y - 15 = 0
4) 5x² + 5y² - 10x + 20y - 30 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(-10)/2(5) = 1
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(20)/2(5) = -2
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (1)² + (-2)² - (-30/5) = 1 + 4 + 6 = 11
- We divide F by 5 because the coefficient of x² and y²
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x - 1)² + (y + 2)² = 11 ⇒ there is no choice
5) 2x² + 2y² - 24x - 16y - 8 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(-24)/2(2) = 6
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(-16)/2(2) = 4
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (6)² + (4)² - (-8/2) = 36 + 16 + 4 = 56
- We divide F by 2 because the coefficient of x² and y²
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x - 6)² + (y - 4)² = 56 ⇒ 2x² + 2y² - 24x - 16y - 8 = 0
6) x² + y² + 2x - 12y - 9 = 0
- we will start to find h and k
∵ h = -coefficient x ÷ 2 coefficient x²
∴ h = -(2)/2(1) = -1
∵ k = -coefficient y ÷ 2 coefficient y²
∴ k = -(-12)/2(1) = 6
∵ r² = h² + k² - F
- where F is the numerical term of the general form
∴ r² = (-1)² + (6)² - (-9) = 1 + 36 + 9 = 46
∴ The equation of the circle in standard form is:
* (x - h)² + (y + k)² = r²
∴ (x + 1)² + (y - 6)² = 46 ⇒ x² + y² + 2x - 6y - 9 = 0
