Question

7.find the equation of a circle with one of the diameters x+2y+8=0and passing through the intersection of a circle x²+y²+2=0 and x²+y²-3y=0.

ans:

`\bold=4x²+4y²+22x+21y=0.}`​

Answer 1

Answer is in the picture

Answer 2

• See below

Explanation:

Find the intersection of circles

• x² + y² + 2x = 0 and
• x² + y² - 3y = 0

Subtract the equations to find:

• 2x = -3y ⇒ y = -2/3x

By substitution find the points:

• (0, 0) and (-1.385, 0.923)

Find midpoint between these two points:

• x = (0- 1.385)/2 = -0.6925 , y = (0 + 0.923)/2 = 0.4615

Find perpendicular bisector passing through this point:

• y - 0.4615 = 3/2(x + 0.6925)
• y = 1.5x + 0.4615 + 1.5(0.6925)
• y = 1.5x + 1.50025

Find intersection of the lines, it is the center:

• y = 1.5x + 1.50025
• x + 2y + 8 = 0

Solving we get:

• x = -2,75, y = -2.625

Find radius, the distance from center to point (0, 0) and get equation of circle:

• (x + 2.75)² + (y + 2.625)² = (0 + 2.75)² + (0 + 2.625)²
• x² + 5.5x + y² + 5.25y = 0
• 4x² + 22x + y² + 21y = 0
• 4x² + 4y² + 22x + 21y = 0