Question

Let the mirror image of a circle c₁ :x² + y² – 2x – 6y + α = 0 in line y = x + 1 be c₂ : 5x² + 5y² + 10gx + 10fy + 38 = 0. If r is the radius of circle c₂, then α + 6r² is equal to _________.

Answer 1

Correct Answer: 12

To solve the problem, we need to find the mirror image of the circle c₁: x² + y² – 2x – 6y + α = 0 in the line y = x + 1 and compare it to the circle c₂: 5x² + 5y² + 10gx + 10fy + 38 = 0. First, simplify c₁:

c₁ can be rewritten as (x – 1)² + (y – 3)² = 10 – α, giving it a center at (1, 3) and radius √(10 – α).

Next, find the mirror image of the center (1, 3) across the line y = x + 1. The perpendicular slope is -1 and the intersection point is found using the equation of this perpendicular line y - 3 = -1(x - 1), which simplifies to y = -x + 4. At the intersection, x + 1 = -x + 4, so x = 1.5 and y = 2.5. The mirror image is thus (2, 2).

Circle c₂ is given as 5x² + 5y² + 10gx + 10fy + 38 = 0. Factor to get x² + y² + 2gx + 2fy + (38/5) = 0, which centers the circle at (-g, -f). Set (-g, -f) = (2, 2), so g = -2, f = -2.

The equation for c₂ becomes x² + y² - 4x - 4y + 38/5 = 0, giving a radius of √[ ((-2)² + (-2)²) - 38/5 ] = √[8 - 38/5] = √[2/5].

Now, compute α + 6r²: Radius r = √(2/5), so r² = 2/5. Then, 6r² = 6(2/5) = 12/5. Using the earlier equation for the radius, 10-α = 2/5 leads to α = 48/5. Calculate α + 6r² = 48/5 + 12/5 = 60/5 = 12.

The solution confirms that α + 6r² = 12, which is within the expected range [12,12].