Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0
At y-intercept, 3(0) — 2y + 6 = 0 and solving for y, y = 3.
Therefore, the coordinates of y — intercept is (0, 3).
Then, r\(^2\) = (0 — 2)\(^2\) + (3 — 3)\(^2\) = 4. With the centre at (2, 3). the equation of the circle was (x — 2)\(^2\) + (y — 3)\(^2\) = 4 and which simplified to \(x^2\) - y\(^2\) — 4x — 6y + 9 = 0
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