- A circle has its center on the line 2y=3x and tangent to the x-axis at (4,0). Find its equation.

the center of the circle lies on the line that is normal to the x-axis and goes trough (4, 0) that is the line x = 4

now we know the center lies on the intersection of 2y = 3x and x = 4

by solving

2y = 3x

x = 4

we find

y = 6

thus the coordinates of the center are O(4 , 6)

lets find the radius r. the radius is equal to the distance of O(4 , 6) to the x-axis follows r = 6

the equation of the circle is

(x - 4)^2 + (y - 6)^2 = 36

click here to see the graph of the circle

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- Find the points of intersection of the circles x^2+y^2-18x-4y+35=0 and x^2+y^2+2x+6y-15=0

we find the intersections by solving the system of equations

x^2+y^2-18x-4y+35=0 is equivalent to (x - 9)^2 + (y - 2)^2 = 50

x^2+y^2+2x+6y-15=0 is equivalent to (x + 1)^ + (y + 3)^2 = 25

if we draw them we see the intersection points (2, 1) and (4, -3)

click here to see the graph

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