Mathematics - Circle Question with Solution | TestHub

Comparing this with the general standard form x² + y² + 2gx + 2fy + c = 0:

• Center (O₁): (3, 3)

• Radius (r₁): √(g² + f² - c) = √((-3)² + (-3)² - 14) = √(9 + 9 - 14) = 2

2. Establish Conditions for the Variable Circle

Let the center of the variable circle be P(h, k) and its radius be r.

• Condition 1: The circle touches the y-axis.

The perpendicular distance from the center (h, k) to the y-axis is equal to its radius:

r = |h|

Assuming the circle lies in the region corresponding to the given circle (h > 0), we have r = h.

• Condition 2: The circle touches the given circle externally.

The distance between their centers is equal to the sum of their radii:

O₁P = r + r₁

√((h - 3)² + (k - 3)²) = h + 2

3. Simplify to Find the Locus

Squaring both sides of the equation:

(h - 3)² + (k - 3)² = (h + 2)²

h² - 6h + 9 + k² - 6k + 9 = h² + 4h + 4

Cancel out h² from both sides and collect all terms:

k² - 6k - 6h - 4h + 18 - 4 = 0

k² - 6k - 10h + 14 = 0

Replace (h, k) with (x, y) to write the final equation:

y² - 10x - 6y + 14 = 0 (which represents a parabola)