The equation of the circle which passes through the point (4,5) and its centre at (2,2) is Option 1: (x-2)+(y-2)=13 Option 2: (x-2)^2+(y-2)^2=13 Option 3: x^2+y^2=13 Option 4: (x-4)^2+(y-5)^2=13

The answer is '(x-2)^2+(y-2)^2=13'. View full solution:

JIPMAT - The equation of the circle which passes through the point (4,5) and its centre at (2,2) is | PYQs + Solutions

JIPMAT

Geometry

Medium

$(x-2)+(y-2)=13$

$(x-2)^{2}+(y-2)^{2}=13$

$x^{2}+y^{2}=13$

$(x-4)^{2}+(y-5)^{2}=13$

✅ Correct Option: 2

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