the image of the circle x2+y2-2x+3y=8 after being reflected by the line y=2x-3 is. pls anyone

The image of the circle is 9·x² + 9·y² - 42·x + 33·y = 22

Step-by-step explanation:

We convert the equation to the form (x - h)² + (y - k)² = r²

Where:

(h, k) = Coordinates of the center of the circle

We therefore have;

x² + y² - 2·x + 3·y = 8 which gives;

x² - 2·x + 1 - 1 + y² + 3·y + (3/2)² - (3/2)² = 8

(x - 1)² + (y + 3/2)² - 1 - (3/2)² = 8

(x - 1)² + (y + 3/2)²  = 8 +  1 + (3/2)² = 45/4

Therefore, the center = (1, -3/2) and the radius = √5 × 3/2

The line joining the centers of the actual (1, -3/2) and reflected circle (h, k) is bisected by the line of reflection, therefore the midpoint of the line is on the line of reflection with coordinates;

Midpoint between (1, -3/2) and  (h, k) =

Which gives;

y = 2·x - 3 being equivalent to

Which gives;

4·h -2·k -13 = 0.......(1)

Similarly the slope of the reflecting line is perpendicular to the line joining the centers of the actual and reflected circles

Therefore;

The slope of the line, y = 2·x - 3 = 2, hence we have;

Which gives;

2·h -4·k -12 = 0.......(2)

Multiplying equation (2) and subtracting from equation (1) gives

4·h -2·k -13 - 2×(2·h -4·k -12) = 0

6·k + 11 = 0

k = -11/6

From which we have;

2·h -4×(-11/6) -12 = 0

Hence;

h = 7/3

Therefore, the center of the reflected circle = (7/3, -11/6)

The equation of the reflected circle is therefore;

(x - 7/3)² + (y + 11/6)² = 45/4

Which gives;

9·x² + 9·y² - 42·x + 33·y - 22 = 0

or x² + y² - 42/9·x + 33/9·y = 22/9.