Question

Number of circles intersecting \[ x^2+y^2=4,\quad x^2+y^2-2x-3=0 \] and \[ x^2+y^2-2y-3=0 \] orthogonally is:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(\infty\)

Hint:

For two circles to cut orthogonally, use \(2gg_1+2ff_1=c+c_1\). After finding the circle, always check whether its radius squared is non-negative.

Answer 1

The Correct Option is A

Step 1: Write the general circle equation.
Let the required circle be x² + y² + 2gx + 2fy + c = 0. For orthogonal intersection with another circle x² + y² + 2g₁x + 2f₁y + c₁ = 0, the condition is 2gg₁ + 2ff₁ = c + c₁.

Step 2: Apply orthogonality with x² + y² = 4.
Here g₁ = 0, f₁ = 0, c₁ = -4. The condition gives 0 = c - 4, so c = 4.

Step 3: Apply orthogonality with x² + y² - 2x - 3 = 0.
Here g₁ = -1, f₁ = 0, c₁ = -3. The condition gives 2g(-1) + 0 = c - 3 → -2g = 4 - 3 = 1 → g = -1/2.

Step 4: Apply orthogonality with x² + y² - 2y - 3 = 0.
Here g₁ = 0, f₁ = -1, c₁ = -3. The condition gives 0 + 2f(-1) = c - 3 → -2f = 1 → f = -1/2.

Step 5: Check whether the resulting circle is real.
The circle is x² + y² - x - y + 4 = 0. Its radius squared is r² = g² + f² - c = (1/4) + (1/4) - 4 = 1/2 - 4 = -7/2. Since r²<0, no real circle exists.

Step 6: Final conclusion.
The number of real circles satisfying all conditions is 0.