Question

If the circle S = 0 cuts the circles x² + y² - 2x + 6y = 0, x² + y² - 4x - 2y + 6 = 0, and x² + y² - 12x + 2y + 3 = 0 orthogonally, then the equation of the tangent at (0, 3) on S = 0 is:

• A. x + y - 3 = 0
• B. \(y = 3\)
• C. \(x = 0\)
• D. \(x - y + 3 = 0\)

Hint:

Remember the condition for orthogonality of circles and the equation of the tangent to a circle.

Answer 1

The Correct Option is B

Step 1: Let the equation of the circle S be x² + y² + 2gx + 2fy + c = 0.
Since S cuts the given circles orthogonally, we have:
 

1. 2g(-1) + 2f(3) = c + 0 ⇒ -2g + 6f = c ...(1)
2. 2g(-2) + 2f(-1) = c + 6 ⇒ -4g - 2f = c + 6 ...(2)
3. 2g(-6) + 2f(1) = c + 3 ⇒ -12g + 2f = c + 3 ...(3)

Step 2: Solve the equations (1), (2), and (3).
From (1) and (2):
-2g + 6f = c
-4g - 2f = c + 6
Subtracting the equations:
2g + 8f = -6 ⇒ g + 4f = -3 ⇒ g = -3 - 4f ...(4)
From (1) and (3):
-2g + 6f = c
-12g + 2f = c + 3
Subtracting the equations:
10g + 4f = -3 ...(5)
Substitute (4) in (5):
10(-3 - 4f) + 4f = -3
-30 - 40f + 4f = -3
-36f = 27
f = -\(\frac{27}{36}\) = -\(\frac{3}{4}\)
Substitute f = -\(\frac{3}{4}\) in (4):
g = -3 - 4(-\(\frac{3}{4}\)) = -3 + 3 = 0
Substitute g = 0 and f = -\(\frac{3}{4}\) in (1):
c = -2(0) + 6(-\(\frac{3}{4}\)) = -\(\frac{9}{2}\)
So, the equation of circle S is x² + y² - \(\frac{3}{2}\)y - \(\frac{9}{2}\) = 0.
2x² + 2y² - 3y - 9 = 0

Step 3: Find the equation of the tangent at (0, 3).
The equation of the tangent at (x₁, y₁) to x² + y² + 2gx + 2fy + c = 0 is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
Here, (x₁, y₁) = (0, 3), g = 0, f = -\(\frac{3}{4}\), c = -\(\frac{9}{2}\).
x(0) + y(3) + 0(x + 0) - \(\frac{3}{4}\)(y + 3) - \(\frac{9}{2}\) = 0
3y - \(\frac{3}{4}\)y - \(\frac{9}{4}\) - \(\frac{18}{4}\) = 0
12y - 3y - 9 - 18 = 0
9y - 27 = 0
9y = 27
y = 3
Therefore, the equation of the tangent at (0, 3) on S = 0 is y = 3.