If the line y=x+3 meets the circle x^2 + y^2 = a^2 at A and B, then equation of the circle on AB as diametre is...

Question

If the line y=x+3 meets the circle x^2 + y^2 = a^2 at A and B, then equation of the circle on AB as diameter is...

Answer 1

AB is the diameter of the new circle with center at the midpoint of AB.

To find the x-coordinates of A and B respectively, substitute x+3 for y in the equation of the circle:

x² + (x+3)² = a²

x² + x² + 6x + 9 = a²

2x² + 6x + (9 - a²) = 0

Using the quadratic formula and simplifying:

x₁ = [(-3 + √(2a²-9))/2]

x₂ = [(-3 - √(2a²-9))/2]

This leads to the conclusion that a must be greater than 3√2/2, i.e. discriminant must be positive.

The y coordinates corresponding to x₁ and x₂, can be found from the equation of the line, i.e. just add 3.

y₁ = x₁ + 3

y₂ = x₂ + 3

A= (x₁, y₁)

B= (x₂, y₂)

For a circle with AB as a diameter, the center of that circle is located at the midpoint of AB.

To get the x-coordinate of the midpoint: (x₁+x₂)/2 = (-6/2)/2 = -3/2

To get the y-coordinate of the midpoint: (y₁+y₂)/2 = 3/2

The center of the circle with diameter on AB is always M(-3/2, 3/2) regardless of the value of "a". Cool!

We still need an expression for the radius of the circle, which will require the distance from M to A or M to B.

Since the equation of that circle will be R², just find (x₁ + 3/2)² + (y₁ - 3/2)². The work is not shown here--left to you--but the result for R² is (2a² - 9)/2.

So the equation of the circle with diameter on segment AB is:

(x + 3/2)² + (y - 3/2)² = (2a²- 9)/2

If you square the binomials, multiply every term by 4 to clear fractions, set equal to zero, you will notice every term is divisible by 4, leaving:

x² + 3x + y² - 3y + 9 - a² = 0

Visit this graph and use the slider on "a" to see that the midpoint of segment AB is always (-1.5, 1.5).

desmos.com/calculator/kihtdfhqeu