Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0

Assessment: WAEC SSCE - Further Mathematics - 2019 (Essay) Subject: Further Mathematics

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First find the y-intercept of \(3x-2y+6=0\): put \(x=0\Rightarrow -2y+6=0\Rightarrow y=3\). The point is \((0,3)\).

The radius is the distance from the centre \((2,3)\) to \((0,3)\):

\[r=\sqrt{(2-0)^2+(3-3)^2}=\sqrt{4}=2.\]

Equation of the circle:

\[(x-2)^2+(y-3)^2=2^2\Rightarrow (x-2)^2+(y-3)^2=4,\]

or \(x^2+y^2-4x-6y+9=0\).

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