Question

The line $y = x$ touches a circle at a point $P$ in the first quadrant such that the distance of $P$ from the origin is $5\sqrt{2}$. If the length of the portion intercepted by this circle on the y-axis is 5 and the equation of the circle, is $x^2 + y^2 + Ax + By + C = 0$. Then find the value of $\frac{|A+B+C|}{6}$.

Answer 1

5

Answer 2

Here's how to solve the problem:

1. Find the coordinates of point P:

   Since $P$ lies on the line $y = x$ and is at a distance of $5\sqrt{2}$ from the origin, its coordinates are $(5, 5)$. This is because the distance from the origin to $(p, p)$ is $\sqrt{p^2 + p^2} = p\sqrt{2}$, so $p\sqrt{2} = 5\sqrt{2}$ implies $p = 5$.

2. Use the circle equation and point P:

   Since $P(5, 5)$ lies on the circle $x^2 + y^2 + Ax + By + C = 0$, we have:

   $5^2 + 5^2 + 5A + 5B + C = 0$ $5A + 5B + C = -50$ (Equation 1)

3. Use the tangent condition:

   The line $y = x$ is tangent to the circle at $P(5, 5)$. The radius from the center $Q(-\frac{A}{2}, -\frac{B}{2})$ to the point $P$ is perpendicular to the tangent line. The slope of the tangent line is 1. The slope of the radius $QP$ is:

   $m_{QP} = \frac{5 + \frac{B}{2}}{5 + \frac{A}{2}} = \frac{10 + B}{10 + A}$

   Since $QP$ is perpendicular to the tangent line, the product of their slopes is -1:

   $\frac{10 + B}{10 + A} \cdot 1 = -1$ $10 + B = -10 - A$ $A + B = -20$ (Equation 2)

4. Use the y-intercept condition:

   The length of the intercept on the y-axis is 5. Setting $x = 0$ in the circle equation gives $y^2 + By + C = 0$. The difference between the roots $y_1$ and $y_2$ of this quadratic is the length of the intercept, so $|y_1 - y_2| = 5$. Using the formula for the difference of roots, we have:

   $\sqrt{B^2 - 4C} = 5$ $B^2 - 4C = 25$ (Equation 3)

5. Solve the system of equations:

   We have the following system of equations:

   • $5A + 5B + C = -50$
   • $A + B = -20$
   • $B^2 - 4C = 25$

   Substitute Equation 2 into Equation 1:

   $5(-20) + C = -50$ $C = 50$

   Substitute $C = 50$ into Equation 3:

   $B^2 - 4(50) = 25$ $B^2 = 225$ $B = \pm 15$

   If $B = 15$, then $A = -35$. If $B = -15$, then $A = -5$.

   So we have two possible sets of values: $(A, B, C) = (-35, 15, 50)$ and $(-5, -15, 50)$.

6. Calculate the final value:

   In both cases, $A + B + C = -20 + 50 = 30$. Therefore, $\frac{|A + B + C|}{6} = \frac{30}{6} = 5$.

Therefore, the final answer is 5.