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Answer:

To find the limiting point of the coaxial system of two circles represented by the equations:

1. \(x² + y² - 8x + 2y + 32 = 0\)

2. \(x² + y² - 7x + 23 = 0\)

We can use the concept of radical center. The radical center is the point from which the power of a circle is the same for both circles.

The general equation of a circle is \(x² + y² + 2gx + 2fy + c = 0\), where the center of the circle is \((-g, -f)\) and the radius is \(r = \sqrt{g² + f² - c}\).

For the first circle (equation 1):

- Center: \((4, -1)\)

- Radius: \(r₁ = \sqrt{4² + (-1)² - 32} = \sqrt{16 + 1 - 32} = \sqrt{-15}\) (since the radius is imaginary, this circle doesn't intersect the real plane)

For the second circle (equation 2):

- Center: \((\frac{7}{2}, -\frac{23}{2})\)

- Radius: \(r₂ = \sqrt{\left(\frac{7}{2}\right)² + \left(-\frac{23}{2}\right)² - 23} = \sqrt{\frac{49}{4} + \frac{529}{4} - 23} = \sqrt{\frac{578 - 92}{4}} = \sqrt{\frac{486}{4}} = \sqrt{121.5} = 11\sqrt{\frac{9}{4}} = \frac{33}{2}\)

Now, we need to find the radical center, which is the point \((x, y)\) from which the power of both circles is the same. The power of a point with coordinates \((x, y)\) with respect to a circle with center \((h, k)\) and radius \(r\) is given by \(P = (x-h)² + (y-k)² - r²\).

Let's set up equations for the power of the point \((x, y)\) with respect to both circles:

1. For the first circle:

\(P₁ = (x - 4)² + (y + 1)² - (-15)\)

2. For the second circle:

\(P₂ = \left(x - \frac{7}{2}\right)² + \left(y + \frac{23}{2}\right)² - \left(\frac{33}{2}\right)²\)

Now, since the point \((x, y)\) is the radical center, \(P₁\) and \(P₂\) must be equal:

\(P₁ = P₂\)

Substitute the expressions for \(P₁\) and \(P₂\):

\((x - 4)² + (y + 1)² - (-15) = \left(x - \frac{7}{2}\right)² + \left(y + \frac{23}{2}\right)² - \left(\frac{33}{2}\right)²\)

Expand and simplify:

\((x - 4)² + (y + 1)² + 15 = \left(x - \frac{7}{2}\right)² + \left(y + \frac{23}{2}\right)² + \left(\frac{33}{2}\right)²\)

Now, let's solve for \(x\) and \(y\):

\((x - 4)² - \left(x - \frac{7}{2}\right)² + (y + 1)² - \left(y + \frac{23}{2}\right)² = \left(\frac{33}{2}\right)² - 15\)

Use the difference of squares:

\((x - 4 + x - \frac{7}{2})(x - 4 - x + \frac{7}{2}) + (y + 1 + y + \frac{23}{2})(y + 1 - y - \frac{23}{2}) = \left(\frac{33}{2}\right)² - 15\)

Simplify:

\(\left(\frac{3x}{2} - \frac{15}{2}\right)\left(-\frac{3x}{2} + \frac{15}{2}\right) + \left(2y + \frac{25}{2}\right)\left(-\frac{25}{2}\right) = \left(\frac{33}{2}\right)² - 15\)

Now, solve for \(x\) and \(y\):

\(\frac{9x²}{4} - \frac{225}{4} - \frac{625}{4} = \frac{1089}{4} - 15\)

Combine like terms:

\(\frac{9x²}{4} - \frac{850}{4} = \frac{1089}{4} - 15\)

\(\frac{9x²}{4} - \frac{850}{4} = \frac{1089 - 60}{4}\)

\(\frac{9x²}{4} - \frac{850}{4} = \frac{1029}{4}\)

Now, isolate \(x²\):

\(\frac{9x²}{4} = \frac{850}{4} + \frac{1029}{4}\)

\(\frac{9x²}{4} = \frac{1879}{4}\)

Now, solve for \(x²\):

\(9x² = 1879\)

Divide by 9:

\(x² = \frac{1879}{9}\)

Now, take the square root of both sides:

\(x = \sqrt{\frac{1879}{9}}\)

Simplify:

\(x = \frac{\sqrt{1879}}{3}\)

So, the x-coordinate of the limiting point is \(x = \frac{\sqrt{1879}}{3}\).

Now, let's solve for \(y\) using one of the equations:

\((x - 4)² + (y + 1)² - (-15) = 0\)

Plug in the value of \(x\):

\(\left(\frac{\sqrt{1879}}{3} - 4\right)² + (y + 1)² + 15 = 0\)

Now, solve for \(y\):

\(\left(\frac{\sqrt{1879}}{3} - 4\right)² + (y + 1)² + 15 - 15 = 0 - 15\)

\(\left(\frac{\sqrt{1879}}{3} - 4\right)² + (y + 1)² = -15\)

Since squares are always non-negative, there are no real solutions for \(y\). Therefore, the limiting point of this coaxial system does not exist in the real plane. It may be an imaginary point depending on the context of the