Find the equation of a circle with centre at (2, - 3) and radius 5 uni

1.	Find the equation of a circle with centre at (2, - 3) and radius 5 units.1. x2 + y2 - 4x + 6y - 12 = 02. x2 + y2 - 4x - 6y - 12 = 03. x2 + y2 + 10x + 2y + 22 = 04. None of these
Answer» Correct Answer - Option 1 : x² + y² - 4x + 6y - 12 = 0 Concept: The equation of circle with centre at (h, k) and radius 'r' is (x - h)2 + (y - k)2 = r2 Calculation: We know that, the equation of circle with centre at (h, k) and radius 'r' is (x - h)²+ (y - k)²= r² Here, centre (h, k) = (2, - 3) and radius r = 5 units. Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is (x - 2)2 + (y + 3)2 = 52 ⇒\(\rm x^2 - 4x + 4 + y^2 +6y +9 = 25\) ⇒\(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \) Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \).

Find the equation of a circle with centre at (2, - 3) and radius 5 units.1. x2 + y2 - 4x + 6y - 12 = 02. x2 + y2 - 4x - 6y - 12 = 03. x2 + y2 + 10x + 2y + 22 = 04. None of these

Answer» Correct Answer - Option 1 : x² + y² - 4x + 6y - 12 = 0

Concept:

The equation of circle with centre at (h, k) and radius 'r' is

(x - h)2 + (y - k)2 = r2

Calculation:

We know that, the equation of circle with centre at (h, k) and radius 'r' is

(x - h)²+ (y - k)²= r²

Here, centre (h, k) = (2, - 3) and radius r = 5 units.

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is

(x - 2)2 + (y + 3)2 = 52

⇒\(\rm x^2 - 4x + 4 + y^2 +6y +9 = 25\)

⇒\(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \)

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \).

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