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Problem 2: Determining the center \(C\) and radius \(R\) of a circle

Given a circle that passes through the point \(A( 4, 2) \) and is tangent to both coordinate axes, we need to find its center \(C\) and radius \(R\).

For a circle to be tangent to both the x-axis and y-axis, its center \(C\) must be at \(( R, R) \), where \(R\) is the radius of the circle.

Given that the circle passes through the point \(A( 4, 2) \), the distance from the center \(( R, R) \) to the point \(A( 4, 2) \) is equal to the radius \(R\). Therefore, we can use the distance formula:

\[\sqrt { ( R - 4) ^ 2 + ( R - 2) ^ 2} = R\]

Squaring both sides to eliminate the square root:

\[( R - 4) ^ 2 + ( R - 2) ^ 2 = R^ 2\]

Expanding and simplifying:

\[( R^ 2 - 8R + 16) + ( R^ 2 - 4R + 4) = R^ 2\]

\[2R^ 2 - 12R + 20 = R^ 2\]

\[R^ 2 - 12R + 20 = 0\]

Solving this quadratic equation:

\[R = \frac { 12 \pm \sqrt { 144 - 80} } { 2} \]

\[R = \frac { 12 \pm \sqrt { 64} } { 2} \]

\[R = \frac { 12 \pm 8} { 2} \]

\[R = 10 \quad \text { or} \quad R = 2\]

Since the point \(A( 4, 2) \) is relatively close, we choose \(R = 2\).

So, the radius \(R = 2\), and the center \(C = ( 2, 2) \);  the radius \(R = 10\), and the center \(C = ( 10, 10) \).

Problem 3: Finding the coordinates of point \(P\)

Given that the distance of point \(P\) to the x-axis is \(\frac { 5} { 3} \) times its distance to the y-axis, and the area of the triangle formed by points \(P\), \(A( - 2, 3) \), and \(B( 4, 5) \) is \(15 \space \text { u} ^ 2\).

We have two possible scenarios based on the distance ratio:

1. \(y = \frac { 5} { 3} x\)

2. \(y = - \frac { 5} { 3} x\)

Case 1: \(y = \frac { 5} { 3} x\)

Using the area formula for a triangle:

\[\text { Area} = \frac { 1} { 2} \left | x_ 1( y_ 2 - y_ 3) + x_ 2( y_ 3 - y_ 1) + x_ 3( y_ 1 - y_ 2) \right | \]

Substituting points \(A( - 2, 3) \), \(B( 4, 5) \), and \(P( x, \frac { 5} { 3} x) \):

\[15 = \frac { 1} { 2} \left | - 2( 5 - \frac { 5} { 3} x) + 4( \frac { 5} { 3} x - 3) + x( 3 - 5) \right | \]

Simplifying:

\[30 = \left | - 2( 5 - \frac { 5} { 3} x) + 4( \frac { 5} { 3} x - 3) + x( 3 - 5) \right | \]

\[30 = \left | - 10 + \frac { 10} { 3} x + \frac { 20} { 3} x - 12 - 2x \right | \]

\[30 = \left | - 22 + \frac { 30} { 3} x - 2x \right | \]

\[30 = \left | - 22 + 10x - 2x \right | \]

\[30 = \left | - 22 + 8x \right | \]

This gives us two equations:

1. \(- 22 + 8x = 30\)

2. \(- 22 + 8x = - 30\)

Solving the first equation:

\[- 22 + 8x = 30\]

\[8x = 52\]

\[x = 6.5\]

\[y = \frac { 5} { 3} \times 6.5 = \frac { 32.5} { 3} \approx 10.83\]

So one possible point \(P\) is \(( 6.5, 10.83) \).

Solving the second equation:

\[- 22 + 8x = - 30\]

\[8x = - 8\]

\[x = - 1\]

\[y = \frac { 5} { 3} \times - 1 = - \frac { 5} { 3} \approx - 1.67\]

So another possible point \(P\) is \(( - 1, - 1.67) \).

Case 2: \(y = - \frac { 5} { 3} x\)

Substituting points \(A( - 2, 3) \), \(B( 4, 5) \), and \(P( x, - \frac { 5} { 3} x) \):

\[15 = \frac { 1} { 2} \left | - 2( 5 - ( - \frac { 5} { 3} x) ) + 4( - \frac { 5} { 3} x - 3) + x( 3 - 5) \right | \]

Simplifying:

\[30 = \left | - 2( 5 + \frac { 5} { 3} x) + 4( - \frac { 5} { 3} x - 3) + x( 3 - 5) \right | \]

\[30 = \left | - 10 - \frac { 10} { 3} x - \frac { 20} { 3} x - 12 - 2x \right | \]

\[30 = \left | - 22 - \frac { 30} { 3} x - 2x \right | \]

\[30 = \left | - 22 - 10x - 2x \right | \]

\[30 = \left | - 22 - 12x \right | \]

This gives us two equations:

1. \(- 22 - 12x = 30\)

2. \(- 22 - 12x = - 30\)

Solving the first equation:

\[- 22 - 12x = 30\]

\[- 12x = 52\]

\[x = - \frac { 52} { 12} = - \frac { 26} { 6} = - 4.33\]

\[y = - \frac { 5} { 3} \times - 4.33 = \frac { 21.65} { 3} \approx 7.22\]

So another possible point \(P\) is \(( - 4.33, 7.22) \).

Solving the second equation:

\[- 22 - 12x = - 30\]

\[- 12x = - 8\]

\[x = \frac { 8} { 12} = \frac { 2} { 3} \]

\[y = - \frac { 5} { 3} \times \frac { 2} { 3} = - \frac { 10} { 9} \approx - 1.11\]

So another possible point \(P\) is \(( \frac { 2} { 3} , - 1.11) \).

Conclusion

After recalculating, the possible points \(P\) are \(( 6.5, 10.83) \) and \(( - 4.33, 7.22) \), as well as additional possible points \(( \frac { 2} { 3} , - 1.11) \) , \(( - 1, - 1.67) \) and \(( - 4.33, 7.22) \).

Supplemental Knowledge

Problem 2: Circle Tangent to Coordinate Axes

For a circle that is tangent to both coordinate axes, the center of the circle is equidistant from both axes. If the center is \(C( h, k) \) and the radius is \(R\), then \(h = R\) and \(k = R\). Given that the circle passes through point \(A( 4, 2) \), we can use this information to find \(R\).

The general equation of a circle with center \(( h, k) \) and radius \(R\) is:

\[( x - h) ^ 2 + ( y - k) ^ 2 = R^ 2\]

Problem 3: Distance to X-axis and Y-axis

For a point \(P( x, y) \), the distance to the X-axis is \(| y| \) and the distance to the Y-axis is \(| x| \). The given relationship is:

\[| y| = \frac { 5} { 3} | x| \]

Additionally, we know that connecting point P with points A(-2, 3) and B(4, 5) forms a triangle with an area of 15 square units. The formula for the area of a triangle given by three points \(( x_ 1, y_ 1) \), \(( x_ 2, y_ 2) \), \(( x_ 3, y_ 3) \) is:

\[\text { Area} = \frac { 1} { 2} \left | x_ 1( y_ 2 - y_ 3) + x_ 2( y_ 3 - y_ 1) + x_ 3( y_ 1 - y_ 2) \right | \]

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