(a) Express \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}\) in partial fractions.
(b) The coordinates of the centre and circumference of a circle are (-2, 5) and 6π units respectively. Find the equation of the circle.
(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)
\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)
\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
Put \(x=1\)
\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)
⇒25=25A
=A=\(\frac{25}{25}=1\)
Put \(x=-\frac{3}{2}\)
\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)
⇒15=-2.5C
\(=C=-\frac{15}{2.5}=-6\)
Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)
\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)
\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)
\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)
By comparing the coefficient of \(x, 8=12A+B+c\)
=8=12(1)+B-6
=8=12+B-6
=8=6+B
=8-6=B
=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)
(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)
Where "a" and "b" are the coordinate of the center and "r" is the radius
2πr = 6π (given)
∴ r = 3 units
=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)
= \((x + 2)^2 + (y - 5)^2 = 9\)
= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)
= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)
∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)
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