Question

The equation of a line inclined at an angle $\frac{\pi}{4}$ to the axis X, such that the two circles $x^2+y^2=4, x^2+y^2-10x+14y+65=0$ intercept equal lengths on it, is

• A. 2x-2y-3=0
• B. 2x-2y+3=0
• C. x-y+6=0
• D. x-y-6=0

Answer 1

Answer: (D)

x-y-6=0

Answer 2

The equation of a line inclined at an angle $\theta$ to the X-axis has slope $m = \tan \theta$. Given $\theta = \frac{\pi}{4}$, the slope is $m = \tan(\frac{\pi}{4}) = 1$. The general equation of the line is $y = mx + c$, which becomes $y = x + c$, or $x - y + c = 0$.

The first circle is $C_1: x^2 + y^2 = 4$. Its center is $O_1 = (0,0)$ and its radius is $r_1 = 2$. The second circle is $C_2: x^2 + y^2 - 10x + 14y + 65 = 0$. Completing the square, we get $(x-5)^2 + (y+7)^2 = 25 + 49 - 65 = 9$. Its center is $O_2 = (5,-7)$ and its radius is $r_2 = 3$.

The length of the intercept made by a circle with radius $r$ on a line is given by $2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from the center of the circle to the line. For a real intercept, $r^2 - d^2 \ge 0$.

The perpendicular distance $d_1$ from $O_1(0,0)$ to the line $x - y + c = 0$ is: $d_1 = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$. The square of the intercept length on $C_1$ is proportional to $r_1^2 - d_1^2 = 2^2 - \left(\frac{|c|}{\sqrt{2}}\right)^2 = 4 - \frac{c^2}{2}$.

The perpendicular distance $d_2$ from $O_2(5,-7)$ to the line $x - y + c = 0$ is: $d_2 = \frac{|5 - (-7) + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|12 + c|}{\sqrt{2}}$. The square of the intercept length on $C_2$ is proportional to $r_2^2 - d_2^2 = 3^2 - \left(\frac{|12 + c|}{\sqrt{2}}\right)^2 = 9 - \frac{(12 + c)^2}{2}$.

Since the two circles intercept equal lengths on the line, the squares of these lengths must be equal: $r_1^2 - d_1^2 = r_2^2 - d_2^2$ $4 - \frac{c^2}{2} = 9 - \frac{(12 + c)^2}{2}$

Multiplying the entire equation by 2: $8 - c^2 = 18 - (12 + c)^2$ $8 - c^2 = 18 - (144 + 24c + c^2)$ $8 - c^2 = 18 - 144 - 24c - c^2$ $8 = -126 - 24c$ $24c = -126 - 8$ $24c = -134$ $c = -\frac{134}{24} = -\frac{67}{12}$.

The equation of the line is $x - y - \frac{67}{12} = 0$, or $12x - 12y - 67 = 0$.

However, this derived line does not match any of the options. Let's re-examine the problem and options. It's possible there's an error in the question or options. If we check the condition for real intercepts with $c = -67/12$: For $C_1$: $4 - \frac{c^2}{2} = 4 - \frac{(-67/12)^2}{2} = 4 - \frac{4489}{288} < 0$. This means the line does not intersect the first circle.

Let's assume there is a typo in the radius of the first circle, and it should be $r_1=3$. Then the condition becomes: $9 - \frac{c^2}{2} = 9 - \frac{(12 + c)^2}{2}$ $-\frac{c^2}{2} = -\frac{(12 + c)^2}{2}$ $c^2 = (12 + c)^2$ This implies $c = 12 + c$ (which gives $0=12$, impossible) or $c = -(12 + c)$. $c = -12 - c$ $2c = -12$ $c = -6$.

If $c=-6$, the equation of the line is $x - y - 6 = 0$. This is option (D). Let's check the intercept conditions for this line with the original circles: For $C_1$ ($r_1=2$): $4 - \frac{(-6)^2}{2} = 4 - \frac{36}{2} = 4 - 18 = -14$. (No real intercept) For $C_2$ ($r_2=3$): $9 - \frac{(12 + (-6))^2}{2} = 9 - \frac{6^2}{2} = 9 - \frac{36}{2} = 9 - 18 = -9$. (No real intercept)

Even with the assumption of $r_1=3$, the intercepts are not real. However, if the question implies that the "squared lengths" are equal, then $c=-6$ leads to equal squared lengths of -14 and -9, which are not equal.

There seems to be an error in the question statement or the provided options, as the derived correct line $12x - 12y - 67 = 0$ is not present, and option (D) $x-y-6=0$ does not yield real intercepts for the first circle. However, in some contexts, problems with errors might intend for a specific option to be chosen based on a partial fulfillment of conditions or a common mistake. If we consider the case where the distances from the centers to the line are equal (not the intercept lengths), then: $d_1 = d_2$ $\frac{|c|}{\sqrt{2}} = \frac{|12+c|}{\sqrt{2}}$ $|c| = |12+c|$ $c = 12+c \implies 0=12$ (impossible) $c = -(12+c) \implies c = -12-c \implies 2c = -12 \implies c = -6$. This leads to option (D) $x-y-6=0$. This is a common point of confusion. Given that option (D) is derived from an equality of distances, and often such questions might have errors pointing to related but incorrect conditions, option (D) is the most plausible intended answer despite the mathematical inconsistency with the stated problem of equal intercept lengths.

The line $x-y-6=0$ is inclined at $\frac{\pi}{4}$ to the X-axis and has slope 1.