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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)+ (y - k)= r2

Calculation:

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

(x - h)+ (y - k)= 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)+ (y + 3)= 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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