(3)

We know that Equation of circle is (x - a)² + (y - b)² = r².

Here, (a,b) is the centre of the circle. radius is the distance between the centre of circle and any point on the centre of circle.

Given centre of circle is (2,2) and equation of circle is passing through (4,5)

Here, (x₁,y₁) = (2,2) and (x₂,y₂) = (4,5).

Now,

We know that distance between the two points is calculated as:

⇒ √(x₂ - x₁)² + (y₂ - y₁)²

⇒ √(4 - 2)² + (5 - 2)²

⇒ √(2)² + (3)²

⇒ √4 + 9

⇒ √13.

Substitute the value in (1), we get

⇒ (x - 2)² + (y - 2)² = (√13)²

⇒ x² + 4 - 4x + y² + 4 - 4y = 13

⇒ x² + y² - 4x - 4y + 8 = 13

⇒ x² + y² - 4x - 4y = 5

⇒ x² + y² - 4x - 4y - 5 = 0.

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(14)

Given Equation is x^2 + y^2 - 12x + 4y + 6 = 0.

Option Verification:

(a)

⇒ x + y = 0

⇒ x = -y.

Substitute in (1),we get

⇒ (y)^2 + (y)^2 + 12y + 4y + 6 = 0

⇒ 2y^2 + 16y + 6 = 0

⇒ y^2 + 8y + 3 = 0

⇒ y = √13 - 4, y = -4 - √13

(b)

⇒ x + 3y = 0

⇒ x = -3y.

Substitute in (1),we get

⇒ (-3y)^2 + y^2 - 12(-3y) + 4y + 6 = 0

⇒ 9y^2 + y^2 + 36y + 4y + 6 = 0

⇒ 10y^2 + 40y + 6 = 0

⇒ 5y^2 + 20y + 3 = 0

⇒ y =√85 - 10/5, y = -10 + √85/3

(iii)

x = y

⇒ (y)^2 + (y)^2 - 12(y) + 4y + 6 = 0

⇒ 2y^2 - 12y + 4y + 6 = 0

⇒ 2y^2 - 8y + 6 = 0

⇒ y^2 - 4y + 3 = 0

⇒ y^2 - y - 3y + 3 = 0

⇒ y(y - 1) - 3(y - 1) = 0

⇒ (y - 3)(y - 1) = 0

⇒ y = 3,1.

(iv)

⇒ 3x + 2y = 0

⇒ 3x = -2y

⇒ x = -2y/3

Substitute in (1), we get

⇒ (-2y/3)^2 + y^2 - 12(-2y/3) + 4y + 6 = 0

⇒ (13y^2/9) + 12y + 6 = 0

⇒ y = 3(√246 - 18)/13, -3(18 + √246)/13.

Therefore, the answer is x = y.

Hope this helps!

Verified answer