Question

The absolute difference between the squares of the radii of the two circles passing through the point \( (-9, 4) \) and touching the lines \( x + y = 3 \) and \( x - y = 3 \), is equal to:

Hint:

For problems involving two tangential circles and points of intersection, use the distance formula between the center and line to find the radius.

Answer 1

Correct Answer: 768

Let the center of the first circle be \( (a, 0) \), with radius \( r_1 \). The equation of the circle is: \[ (x - a)^2 + y^2 = r_1^2 \] Now, the distance from the center of the circle to the line \( x + y = 3 \) is the radius \( r_1 \). The distance formula for a point to a line \( Ax + By + C = 0 \) is: \[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Substituting the values, we find the relationship between \( a \) and \( r_1 \). Similarly, for the second circle, we use the equation of the second line \( x - y = 3 \). The result of the calculations is the absolute difference between the squares of the radii: \[ |r_1^2 - r_2^2| = 768 \] 

Answer 2

The problem requires finding the absolute difference between the squares of the radii of two circles. Both circles pass through the point \( P(-9, 4) \) and are tangent to the lines \( L_1: x + y = 3 \) and \( L_2: x - y = 3 \).

Concept Used:

The key concepts for this problem are:

1. The center of a circle that is tangent to two intersecting lines must lie on one of the angle bisectors of those lines.
2. The perpendicular distance from the center of a circle to any tangent line is equal to its radius.
3. The distance from the center of a circle to any point on its circumference is also equal to its radius.

The equation of the angle bisectors of the lines \( A_1x + B_1y + C_1 = 0 \) and \( A_2x + B_2y + C_2 = 0 \) is given by:

\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]

Step-by-Step Solution:

Step 1: Find the equations of the angle bisectors of the tangent lines.

The given tangent lines are \( L_1: x + y - 3 = 0 \) and \( L_2: x - y - 3 = 0 \).

Using the formula for angle bisectors:

\[ \frac{x + y - 3}{\sqrt{1^2 + 1^2}} = \pm \frac{x - y - 3}{\sqrt{1^2 + (-1)^2}} \] \[ \frac{x + y - 3}{\sqrt{2}} = \pm \frac{x - y - 3}{\sqrt{2}} \]

This gives us two bisector equations:

Bisector 1: \( x + y - 3 = x - y - 3 \implies 2y = 0 \implies y = 0 \)

Bisector 2: \( x + y - 3 = -(x - y - 3) \implies x + y - 3 = -x + y + 3 \implies 2x = 6 \implies x = 3 \)

The center of the required circles must lie on either the line \( y = 0 \) or \( x = 3 \).

Step 2: Case 1 - The center of the circle lies on the bisector \( y = 0 \).

Let the center of the circle be \( C(h, 0) \). The radius \( r \) is the perpendicular distance from the center \( C(h, 0) \) to one of the tangent lines, say \( L_1: x + y - 3 = 0 \).

\[ r = \frac{|h + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|h - 3|}{\sqrt{2}} \]

The circle also passes through the point \( P(-9, 4) \). The distance from the center \( C \) to this point must be equal to the radius \( r \). Thus, \( CP^2 = r^2 \).

\[ (h - (-9))^2 + (0 - 4)^2 = \left(\frac{h - 3}{\sqrt{2}}\right)^2 \] \[ (h + 9)^2 + 16 = \frac{(h - 3)^2}{2} \] \[ 2(h^2 + 18h + 81 + 16) = h^2 - 6h + 9 \] \[ 2(h^2 + 18h + 97) = h^2 - 6h + 9 \] \[ 2h^2 + 36h + 194 = h^2 - 6h + 9 \] \[ h^2 + 42h + 185 = 0 \]

Solving this quadratic equation for \( h \):

\[ h = \frac{-42 \pm \sqrt{42^2 - 4(1)(185)}}{2} = \frac{-42 \pm \sqrt{1764 - 740}}{2} = \frac{-42 \pm \sqrt{1024}}{2} = \frac{-42 \pm 32}{2} \]

This gives two possible values for \( h \):

\( h_1 = \frac{-42 + 32}{2} = -5 \)

\( h_2 = \frac{-42 - 32}{2} = -37 \)

These correspond to two different circles.

Step 3: Case 2 - The center of the circle lies on the bisector \( x = 3 \).

Let the center of the circle be \( C(3, k) \). The radius \( r \) is the perpendicular distance from \( C(3, k) \) to \( L_1: x + y - 3 = 0 \).

\[ r = \frac{|3 + k - 3|}{\sqrt{1^2 + 1^2}} = \frac{|k|}{\sqrt{2}} \]

The distance from \( C(3, k) \) to \( P(-9, 4) \) must also be equal to \( r \).

\[ (3 - (-9))^2 + (k - 4)^2 = \left(\frac{k}{\sqrt{2}}\right)^2 \] \[ 12^2 + (k - 4)^2 = \frac{k^2}{2} \] \[ 144 + k^2 - 8k + 16 = \frac{k^2}{2} \] \[ k^2 - 16k + 320 = 0 \]

The discriminant of this quadratic equation is \( D = b^2 - 4ac = (-16)^2 - 4(1)(320) = 256 - 1280 = -1024 \). Since the discriminant is negative, there are no real solutions for \( k \). Therefore, no such circles have their center on the line \( x = 3 \).

Step 4: Calculate the squares of the radii for the two circles from Case 1.

The formula for the radius is \( r = \frac{|h - 3|}{\sqrt{2}} \), so \( r^2 = \frac{(h - 3)^2}{2} \).

For the first circle with \( h_1 = -5 \):

\[ r_1^2 = \frac{(-5 - 3)^2}{2} = \frac{(-8)^2}{2} = \frac{64}{2} = 32 \]

For the second circle with \( h_2 = -37 \):

\[ r_2^2 = \frac{(-37 - 3)^2}{2} = \frac{(-40)^2}{2} = \frac{1600}{2} = 800 \]

Final Computation & Result:

We are asked to find the absolute difference between the squares of the radii of the two circles.

\[ |r_2^2 - r_1^2| = |800 - 32| = 768 \]

The absolute difference between the squares of the radii is 768.