If a circle has its centre on (2, 3) and intersects another circle x2 + y2 + 4x - 8y + 16 = 0 orthogonally, then what will be the equation of the circle?

  1. x2 + y2 + 4x + 6y = 0
  2. x2 + y2 – 4x - 6y – 12 = 0
  3. x2 + y2 – 4x - 6y = 0
  4. x2 + y2 – 4x + 6y - 9 = 0

Answer (Detailed Solution Below)

Option 3 : x2 + y2 – 4x - 6y = 0
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Detailed Solution

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Concept:

Condition of orthogonality = 2g1g2 + 2f1f2 = c1 + c2

Calculation:

Lets equation of that orthogonal circle is x2 + y2 + 2gx + 2fy + c = 0

Given: centre of circle is (2, 3) and it intersects another circle x2 + y2 + 4x - 8y + 16 = 0 orthogonally.

So g1 = - 2, f1 = - 3, g2 = 2, f2 = - 4, c1 = c and c2 = 16

Applying condition of orthogonality with the given circle –

⇒ 2( - 2)(2) + 2.( - 3). ( - 4) = c + 16

⇒ c = 0

So desired equation will be x2 + y2 – 4x - 6y = 0

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