Find the center and radius of the equation of the circle
a). x2 + y2 – 8x +6y + 9 = 0

Compare the given equation with the general equation of a circle

Given equation: x² + y² - 8x +6y +9 = 0

Equation of a circle: x² + y² + 2gx + 2fy + c = 0

2g= -8; g = -8÷2= -4

2f = 6; f = 6÷2 = 3

c= 9

Solving for radius of the circle...

r =√g² + f² - c

g= -4, g² = ±16

f = 3, f² = 9

Substituting, we have

r = √16 + 9 - 9

r = √25-9

r=√16

r = 4

Therefore, the centre and radius of the circle is 9 and 4 respectively.

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the co-ordinates of the transformed rectangle are:

d. (0, 0), (0, 3), (-3, 3), (-3, 0)

step-by-step explanation:

from the columns of the given matrix, the co-ordinates of the rectangle before transformation are:

(0, 0), (0, 3), (3, 3), (3, 0)

r =

transformed matrix can be obtained by multiplying the given matrix by r:

columns of the transformed matrix are:

0   0   -3   -3

0   3     3     0

from the columns of the resulted matrix, we can see that the vertices of the transformed triangle are:

(0, 0), (0, 3), (-3, 3), (-3, 0)

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step-by-step explanation: