Circle with Centre on the Negative x-axis

To find the intercept made by the circle on the y-axis, we will first determine the equation of the circle.

Finding the Equation of the Circle

Let the center of the circle be (a, 0) on the negative x-axis. The radius of the circle is given as 5 units.

Using the distance formula, we can calculate the distance between the center of the circle and the given point (2, 3) as follows:

√[(2 - a)^2 + (3 - 0)^2] = 5

Simplifying the equation, we get:

(2 - a)^2 + 9 = 25

Expanding and rearranging the terms, we have:

a^2 - 4a + 4 + 9 = 25

a^2 - 4a - 12 = 0

Factoring the quadratic equation, we find:

(a - 6)(a + 2) = 0

This gives us two possible values for the x-coordinate of the center of the circle: a = 6 or a = -2.

Calculating the Equation of the Circle

Using the value a = 6, the equation of the circle can be written as:

(x - 6)^2 + y^2 = 25

Using the value a = -2, the equation of the circle becomes:

(x + 2)^2 + y^2 = 25

Intercept Made by the Circle on the y-axis

The intercept made by the circle on the y-axis is the point where the circle intersects the y-axis. This point can be found by setting x = 0 in the equation of the circle.

For the equation (x - 6)^2 + y^2 = 25, substituting x = 0 gives:

(-6)^2 + y^2 = 25

36 + y^2 = 25

y^2 = 25 - 36

y^2 = -11

Since the square of a real number cannot be negative, there are no intercepts on the y-axis for this circle.

For the equation (x + 2)^2 + y^2 = 25, substituting x = 0 gives:

(2)^2 + y^2 = 25

4 + y^2 = 25

y^2 = 25 - 4

y^2 = 21

Again, since the square of a real number cannot be negative, there are no intercepts on the y-axis for this circle either.

Therefore, the circle with a radius of 5 units and its center on the negative x-axis does not intercept the y-axis.