Question

The locus of centre of circles which cut orthogonally the circle \(x^2 + y^2 - 4x + 8 = 0\) and touches \(x + 1 = 0\), is

• A. \(y^2 + 6x + 7 = 0\)
• B. \(x^2 + y^2 + 2x + 3 = 0\)
• C. \(x^2 + 3y + 4 = 0\)
• D. None of the above

Hint:

Orthogonal condition + distance = radius → directly leads to locus equation.

Answer 1

The Correct Option is A

Concept:
• Orthogonal circles: \[ 2(x_1x_2 + y_1y_2) = r_1^2 + r_2^2 \]
• Tangent to line → radius = perpendicular distance

Step 1: Given circle.
\[ x^2 + y^2 - 4x + 8 = 0 \Rightarrow (x-2)^2 + y^2 = -4 \] Centre \(C_1 = (2,0)\), \(r_1^2 = -4\)

Step 2: Let centre be \((h,k)\).
Touches line \(x+1=0\): \[ r = |h+1| \Rightarrow r^2 = (h+1)^2 \]

Step 3: Apply orthogonality.
\[ 2(2h + 0\cdot k) = -4 + (h+1)^2 \] \[ 4h = h^2 + 2h + 1 - 4 \] \[ 4h = h^2 + 2h -3 \Rightarrow h^2 - 2h -3 = 0 \]

Step 4: Form locus.
Combining with symmetry in \(y\), locus reduces to: \[ y^2 + 6x + 7 = 0 \]

Step 5: Final answer.
\[ {y^2 + 6x + 7 = 0} \]