Question

One circle has a radius of 5 and its center at (0, 5). A second circle has a radius of 12 and its centre at (12, 0). The length of a radius of a third circle which passes through the center of the second circle and both points of intersection of the first 2 circles, is equal to:

• A. 13/2
• B. 15/2
• C. 17/2
• D. None of these

Answer 1

Answer: (A)

13/2

Answer 2

Let $C_1$ be the first circle with center $O_1 = (0, 5)$ and radius $r_1 = 5$. Its equation is $x^2 + (y-5)^2 = 5^2$, which simplifies to $x^2 + y^2 - 10y = 0$.

Let $C_2$ be the second circle with center $O_2 = (12, 0)$ and radius $r_2 = 12$. Its equation is $(x-12)^2 + y^2 = 12^2$, which simplifies to $x^2 + y^2 - 24x = 0$.

To find the points of intersection, we subtract the first equation from the second: $(x^2 + y^2 - 24x) - (x^2 + y^2 - 10y) = 0$ $-24x + 10y = 0 \implies y = \frac{12}{5}x$.

Substitute $y = \frac{12}{5}x$ into the equation for $C_1$: $x^2 + (\frac{12}{5}x)^2 - 10(\frac{12}{5}x) = 0$ $x^2 + \frac{144}{25}x^2 - 24x = 0$ Multiplying by 25: $25x^2 + 144x^2 - 600x = 0$ $169x^2 - 600x = 0$ $x(169x - 600) = 0$. The solutions are $x=0$ and $x = \frac{600}{169}$.

If $x=0$, then $y = \frac{12}{5}(0) = 0$. The first intersection point is $(0,0)$. If $x=\frac{600}{169}$, then $y = \frac{12}{5} \times \frac{600}{169} = \frac{1440}{169}$. The second intersection point is $P = (\frac{600}{169}, \frac{1440}{169})$.

The third circle, $C_3$, passes through $O_2(12, 0)$, $(0,0)$, and $P(\frac{600}{169}, \frac{1440}{169})$. Let the equation of $C_3$ be $x^2 + y^2 + 2gx + 2fy + c = 0$.

Since $C_3$ passes through $(0,0)$, we get $c=0$. The equation is $x^2 + y^2 + 2gx + 2fy = 0$. Since $C_3$ passes through $(12,0)$: $12^2 + 0^2 + 2g(12) + 2f(0) = 0 \implies 144 + 24g = 0 \implies g = -6$. The equation is $x^2 + y^2 - 12x + 2fy = 0$.

Since $C_3$ passes through $P(\frac{600}{169}, \frac{1440}{169})$: $(\frac{600}{169})^2 + (\frac{1440}{169})^2 - 12(\frac{600}{169}) + 2f(\frac{1440}{169}) = 0$. We know that $x_P^2 + y_P^2 = (\frac{600}{169})^2 + (\frac{1440}{169})^2 = \frac{360000 + 2073600}{169^2} = \frac{2433600}{169^2}$. A simpler way is to use $x^2+y^2 = x^2(1 + (\frac{12}{5})^2) = x^2(\frac{169}{25})$. For $P$, $x_P^2+y_P^2 = (\frac{600}{169})^2 \frac{169}{25} = \frac{360000}{169 \times 25} = \frac{14400}{169}$. So, $\frac{14400}{169} - 12(\frac{600}{169}) + 2f(\frac{1440}{169}) = 0$. Multiplying by 169: $14400 - 7200 + 2880f = 0$ $7200 + 2880f = 0 \implies f = -\frac{7200}{2880} = -\frac{5}{2}$.

The equation of $C_3$ is $x^2 + y^2 - 12x - 5y = 0$. The radius of $C_3$ is $r_3 = \sqrt{(-g)^2 + (-f)^2 - c} = \sqrt{(-(-6))^2 + (- (-\frac{5}{2}))^2 - 0} = \sqrt{6^2 + (\frac{5}{2})^2} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144+25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2}$.