Explanation

\((x - a)^{2} + (y - b)^{2} = r^{2}\)

\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Using A(3, 2) :

\(9 - 6a + a^{2} + 4 - 4b + b^{2} = r^{2}\)

\(13 - 6a - 4b + a^{2} + b^{2} = r^{2} .... (1)\)

Using B(-1, -2) :

\(1 + 2a + a^{2} + 4 + 4b + b^{2} = r^{2}\)

\(5 + 2a + 4b + a^{2} + b^{2} = r^{2} .... (2)\)

Using C(5, -4) :

\(25 - 10a + a^{2} + 16 + 8b + b^{2} = r^{2}\)

\(41 - 10a + 8b + a^{2} + b^{2} = r^{2} ... (3)\)

(1) - (2) :

\(8 - 8a - 8b = 0 \implies 1 - a - b = 0 .... (4)\)

(3) - (2) :

\(24 - 12a + 12 + 4b = 0 \implies 36 - 12a + 4b = 0 \)

\(\equiv 9 - 3a + b = 0... (5)\)

Solving (4) and (5), we have :

\(a = 2\frac{1}{2} ; b = -1\frac{1}{2}\)

Coordinates of the centre : \(C(2\frac{1}{2}, -1\frac{1}{2})\).

(b) To get the radius, we can make use of any point on the circle as well as the centre.

Using point \(A(3, 2) & C(2\frac{1}{2}, -1\frac{1}{2})\)

Radius = \(\sqrt{(3 - 2\frac{1}{2})^{2} + (2 + 1\frac{1}{2})^{2}}\)

= \(\sqrt{\frac{1}{4} + \frac{49}{4}}\)

= \(\sqrt{\frac{50}{4}}\)

= \(\frac{5\sqrt{2}}{2} cm\)

(c) Equation of the circle :

\((x - 2\frac{1}{2})^{2} + (y + 1\frac{1}{2})^{2} = (\frac{5\sqrt{2}}{2})^{2}\)

\(x^{2} - 11x + \frac{25}{4} + y^{2} + 3y + \frac{9}{4} = \frac{50}{4}\)

\(x^{2} + y^{2} - 11x + 3y + \frac{34}{4} - \frac{50}{4} = 0\)

\(x^{2} + y^{2} - 11x + 3y - 4 = 0\)

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