Question:

Step 1: Identify the properties of the circles and the common chord. Circle $C_1: x^2 + y^2 = 16$ has its center at $O_1 = (0,0)$ and a radius $r_1 = 4$. Circle $C_2$ has a radius $r_2 = 5$. For the common chord to have maximum length, it must pass through the center of the smaller circle. Since $r_1 < r_2$, $C_1$ is the smaller circle. Therefore, the common chord passes through the center of $C_1$, which is the origin $(0,0)$. Step 2: Determine the length and equation of the common chord. Since the common chord passes through the center of $C_1$, its length is equal to the diameter of $C_1$. Length of common chord $L = 2r_1 = 2(4) = 8$. The common chord passes through the origin $(0,0)$ and has a slope of $\frac{3}{4}$. The equation of the common chord is $y - 0 = \frac{3}{4}(x - 0)$, which simplifies to $4y = 3x$, or $3x - 4y = 0$. Step 3: Establish the relationship between the centers of the circles. Let the center of $C_2$ be $O_2 = (h,k)$. The line joining the centers $O_1(0,0)$ and $O_2(h,k)$ must be perpendicular to the common chord. The slope of the common chord is $m = \frac{3}{4}$. The slope of the line $O_1O_2$ is $m_{\perp} = -\frac{1}{m} = -\frac{4}{3}$. Thus, the slope of the line connecting $(0,0)$ and $(h,k)$ is $\frac{k-0}{h-0} = \frac{k}{h}$. So, $\frac{k}{h} = -\frac{4}{3}$, which implies $3k = -4h$, or $4h + 3k = 0$. Step 4: Calculate the distance from the center of $C_2$ to the common chord. The common chord has length $L=8$ within circle $C_2$, which has radius $r_2=5$. Let $d$ be the perpendicular distance from the center $O_2(h,k)$ to the common chord $3x - 4y = 0$. The relationship between chord length $L$, radius $r_2$, and distance $d$ is given by $L = 2\sqrt{r_2^2 - d^2}$. Substituting the known values: $8 = 2\sqrt{5^2 - d^2}$ $4 = \sqrt{25 - d^2}$ Squaring both sides: $16 = 25 - d^2$ $d^2 = 25 - 16$ $d^2 = 9$ Since distance must be positive, $d = 3$. Using the distance formula from a point $(h,k)$ to the line $Ax+By+C=0$: $d = \frac{|Ah+Bk+C|}{\sqrt{A^2+B^2}}$ For the line $3x - 4y = 0$, the distance from $(h,k)$ is: $d = \frac{|3h - 4k|}{\sqrt{3^2 + (-4)^2}} = \frac{|3h - 4k|}{\sqrt{9 + 16}} = \frac{|3h - 4k|}{5}$. Equating this to the calculated distance $d=3$: $\frac{|3h - 4k|}{5} = 3$ $|3h - 4k| = 15$. Step 5: Solve for the coordinates of the center of $C_2$. We have a system of two equations for $(h,k)$: 1. $4h + 3k = 0$ 2. $|3h - 4k| = 15$ From equation (1), we can express $k$ in terms of $h$: $3k = -4h \implies k = -\frac{4}{3}h$. Substitute this expression for $k$ into equation (2): $\left|3h - 4\left(-\frac{4}{3}h\right)\right| = 15$ $\left|3h + \frac{16}{3}h\right| = 15$ To combine the terms inside the absolute value, find a common denominator: $\left|\frac{9h}{3} + \frac{16h}{3}\right| = 15$ $\left|\frac{25h}{3}\right| = 15$ Since $25/3$ is positive, we can write: $\frac{25|h|}{3} = 15$ $|h| = \frac{15 \times 3}{25} = \frac{45}{25} = \frac{9}{5}$. This gives two possible values for $h$: $h = \frac{9}{5}$ or $h = -\frac{9}{5}$. Case 1: If $h = \frac{9}{5}$ Substitute $h = \frac{9}{5}$ into $k = -\frac{4}{3}h$: $k = -\frac{4}{3}\left(\frac{9}{5}\right) = -\frac{36}{15} = -\frac{12}{5}$. So, one possible center for $C_2$ is $\left(\frac{9}{5}, -\frac{12}{5}\right)$. Case 2: If $h = -\frac{9}{5}$ Substitute $h = -\frac{9}{5}$ into $k = -\frac{4}{3}h$: $k = -\frac{4}{3}\left(-\frac{9}{5}\right) = \frac{36}{15} = \frac{12}{5}$. So, the other possible center for $C_2$ is $\left(-\frac{9}{5}, \frac{12}{5}\right)$. The coordinates of the center of $C_2$ are $\left(\frac{9}{5}, -\frac{12}{5}\right)$ and $\left(-\frac{9}{5}, \frac{12}{5}\right)$. These can be compactly written as $\left(\pm\frac{9}{5}, \mp\frac{12}{5}\right)$.