Key Concepts and Formulas

• Intersection of Lines: Solving a system of linear equations to find the point of intersection.
• Tangent to a Circle: A tangent to a circle is perpendicular to the radius at the point of tangency.
• Distance from a Point to a Line: The distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Find the coordinates of point Q.

• What & Why: We are given that line L1 intersects line L2 at point Q. We need to find the coordinates of Q by solving the system of equations formed by the equations of L1 and L2.
• Math: L1: $4x + 3y + 2 = 0$ L2: $3x - 4y - 11 = 0$ Multiply L1 by 4 and L2 by 3: $16x + 12y + 8 = 0$ $9x - 12y - 33 = 0$ Adding these two equations, we get: $25x - 25 = 0$ $x = 1$ Substitute $x = 1$ into L1: $4(1) + 3y + 2 = 0$ $3y = -6$ $y = -2$
• Reasoning: Solving the system of equations gives us the x and y coordinates of the intersection point.

Step 2: Determine the coordinates of the center P of the circle.

• What & Why: We know that the center P lies on line L1: $4x + 3y + 2 = 0$. We also know that L2 is tangent to the circle at Q, which means that the line connecting P and Q is perpendicular to L2. We can use this information to find the coordinates of P.
• Math: Let the coordinates of P be $(h, k)$. Since P lies on L1, we have: $4h + 3k + 2 = 0$ (Equation 1) The slope of L2 is $m_2 = \frac{3}{4}$. Since PQ is perpendicular to L2, the slope of PQ is $m_{PQ} = -\frac{4}{3}$. The slope of PQ can also be expressed as $\frac{k - (-2)}{h - 1} = \frac{k + 2}{h - 1}$. Therefore, $\frac{k + 2}{h - 1} = -\frac{4}{3}$. $3(k + 2) = -4(h - 1)$ $3k + 6 = -4h + 4$ $4h + 3k + 2 = 0$ (Equation 2) This is the same as Equation 1, so we need to use the radius information. The distance between P(h,k) and Q(1,-2) is equal to the radius, which is 5. $\sqrt{(h - 1)^2 + (k + 2)^2} = 5$ $(h - 1)^2 + (k + 2)^2 = 25$ (Equation 3) From Equation 1, $3k = -4h - 2$, so $k = -\frac{4}{3}h - \frac{2}{3}$. Substitute this into Equation 3: $(h - 1)^2 + (-\frac{4}{3}h - \frac{2}{3} + 2)^2 = 25$ $(h - 1)^2 + (-\frac{4}{3}h + \frac{4}{3})^2 = 25$ $(h - 1)^2 + (\frac{4}{3}(1 - h))^2 = 25$ $(h - 1)^2 + \frac{16}{9}(h - 1)^2 = 25$ $\frac{25}{9}(h - 1)^2 = 25$ $(h - 1)^2 = 9$ $h - 1 = \pm 3$ $h = 4$ or $h = -2$ If $h = 4$, then $k = -\frac{4}{3}(4) - \frac{2}{3} = -\frac{18}{3} = -6$. If $h = -2$, then $k = -\frac{4}{3}(-2) - \frac{2}{3} = \frac{6}{3} = 2$. Since the circle lies below the x-axis, the y-coordinate of the center must be negative. Therefore, $P = (4, -6)$.
• Reasoning: We use the fact that the radius is perpendicular to the tangent and the circle lies below the x-axis to determine the correct center.

Step 3: Calculate the distance of P from the line 5x - 12y + 51 = 0.

• What & Why: Now that we have the coordinates of P, we can use the point-to-line distance formula to find the distance from P to the given line.

• Math: The distance from $(4, -6)$ to the line $5x - 12y + 51 = 0$ is: $d = \frac{|5(4) - 12(-6) + 51|}{\sqrt{5^2 + (-12)^2}} = \frac{|20 + 72 + 51|}{\sqrt{25 + 144}} = \frac{|143|}{\sqrt{169}} = \frac{143}{13} = 11$

• Reasoning: Applying the distance formula gives us the desired distance.

Step 4: Re-examine the solution for errors

What & Why: The answer we got from the previous calculations is 11, but the given correct answer is 1. There must be an error in our calculations. Let's check our work. The error is in Step 2. Equation 2 is the same as Equation 1, so we did not get new information from it.

Math: Let's re-examine the given information. L1: $4x + 3y + 2 = 0$ and L2: $3x - 4y - 11 = 0$. The point Q is (1, -2). The center P is (h, k) and lies on L1: $4h + 3k + 2 = 0$. The distance between P and Q is 5, so $(h-1)^2 + (k+2)^2 = 25$. $k = \frac{-4h - 2}{3}$. Substituting: $(h-1)^2 + (\frac{-4h - 2}{3} + 2)^2 = 25$ $(h-1)^2 + (\frac{-4h + 4}{3})^2 = 25$ $(h-1)^2 + \frac{16}{9}(h-1)^2 = 25$ $\frac{25}{9}(h-1)^2 = 25$ $(h-1)^2 = 9$ $h - 1 = \pm 3$ $h = 4$ or $h = -2$ If $h = 4$, then $k = \frac{-4(4) - 2}{3} = \frac{-18}{3} = -6$. So P is (4, -6). If $h = -2$, then $k = \frac{-4(-2) - 2}{3} = \frac{6}{3} = 2$. So P is (-2, 2). Since the circle is below the x-axis, we take P = (4, -6). Let's check the other condition. L2 is tangent to the circle at Q. Slope of L2 is 3/4. Slope of PQ should be -4/3. If P is (4, -6), slope of PQ is $\frac{-6 - (-2)}{4 - 1} = \frac{-4}{3}$. Checks out. If P is (-2, 2), slope of PQ is $\frac{2 - (-2)}{-2 - 1} = \frac{4}{-3}$. Checks out. Since the circle lies below the x-axis, the y-coordinate of the center must be negative. Thus, P = (4, -6). The distance from P to the line $5x - 12y + 51 = 0$ is: $d = \frac{|5(4) - 12(-6) + 51|}{\sqrt{5^2 + (-12)^2}} = \frac{|20 + 72 + 51|}{\sqrt{25 + 144}} = \frac{143}{13} = 11$. Still not 1.

Something is wrong. Let's go back to the beginning.

L1: $4x + 3y + 2 = 0$. L2: $3x - 4y - 11 = 0$. Intersection Q is (1, -2). Let P be (h,k). $4h + 3k + 2 = 0$. The line L2 touches C at Q. PQ is perpendicular to L2. Thus, slope of PQ is -4/3. $\frac{k + 2}{h - 1} = -\frac{4}{3}$. $3k + 6 = -4h + 4$. So, $4h + 3k + 2 = 0$. $(h - 1)^2 + (k + 2)^2 = 25$. The circle lies below the x-axis. $4h + 3k = -2$. $k = \frac{-4h - 2}{3}$. $(h-1)^2 + (\frac{-4h - 2}{3} + 2)^2 = 25$. $(h-1)^2 + (\frac{-4h + 4}{3})^2 = 25$. $(h-1)^2 + \frac{16}{9}(h-1)^2 = 25$. $\frac{25}{9}(h-1)^2 = 25$. $(h-1)^2 = 9$. $h-1 = \pm 3$. $h = 4$ or $h = -2$. If $h = 4$, $k = \frac{-16 - 2}{3} = -6$. P = (4, -6). If $h = -2$, $k = \frac{8 - 2}{3} = 2$. P = (-2, 2). Since the circle is below x-axis, P = (4, -6). However, we should also check the other case. The circle lies below the x-axis. The entire circle must lie below the x-axis. The y-coordinate of Q is -2. Radius is 5. The y-coordinate of the center is -6. -6 + 5 = -1. -6 - 5 = -11. Thus, the entire circle is below the x-axis.

The distance of P(4, -6) from $5x - 12y + 51 = 0$ is $\frac{|20 + 72 + 51|}{\sqrt{25 + 144}} = \frac{143}{13} = 11$.

Let's consider the other option. If the center is (-2, 2), Q is (1, -2). Radius is 5. 2 + 5 = 7. 2 - 5 = -3. Since the circle lies below the x-axis, the y-coordinate must be negative. Hence, P(4, -6). This is the only valid option.

The distance formula is correct. The coordinates of point P are correct.

The given answer must be incorrect. The distance is indeed 11.

I will assume that the line equation was mistyped and the correct equation should be $5x - 12y - 67 = 0$, which gives distance = 1. $\frac{|5(4) - 12(-6) - 67|}{\sqrt{5^2 + (-12)^2}} = \frac{|20 + 72 - 67|}{13} = \frac{25}{13} \neq 1$

The given correct answer is wrong.

Common Mistakes & Tips

• Double-check all calculations to avoid arithmetic errors.
• Remember that the radius is perpendicular to the tangent at the point of tangency.
• Pay close attention to the wording of the problem, such as "below the x-axis," to determine the correct solution.

Summary

We found the intersection point Q of the two lines. Then, we used the fact that the radius is perpendicular to the tangent line and the circle lies below the x-axis to determine the coordinates of the center P of the circle. Finally, we used the point-to-line distance formula to calculate the distance from P to the given line. According to our calculations, the correct distance is 11, but the provided answer is 1.

Final Answer

The final answer is \boxed{11}. Option is not applicable.