Question:

The equation of a circle whose Centre is \((2,-1)\) and which passes through the point \((3,6)\) is

Show Hint

If the circle passes through a point, use the distance formula from the centre to that point to find \(r^2\).
  • \(x^2+y^2+4x+2y-45=0\)
  • \(x^2+y^2-2x+2y-50=0\)
  • \(x^2+y^2+2x+2y-50=0\)
  • \(x^2+y^2-4x+2y-45=0\)
Show Solution
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The Correct Option is D

Solution and Explanation

Concept: The equation of a circle with centre \((h,k)\) is: \[ (x-h)^2+(y-k)^2=r^2 \]

Step 1:
Given centre is: \[ (2,-1) \] So the equation is: \[ (x-2)^2+(y+1)^2=r^2 \]

Step 2:
The circle passes through \((3,6)\).
So radius is the distance between \((2,-1)\) and \((3,6)\). \[ r^2=(3-2)^2+(6+1)^2 \] \[ r^2=1^2+7^2 \] \[ r^2=1+49=50 \]

Step 3:
Substitute \(r^2=50\). \[ (x-2)^2+(y+1)^2=50 \]

Step 4:
Expand. \[ x^2-4x+4+y^2+2y+1=50 \] \[ x^2+y^2-4x+2y+5-50=0 \] \[ x^2+y^2-4x+2y-45=0 \] Therefore, \[ \boxed{x^2+y^2-4x+2y-45=0} \]
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