Consider two circles C_1:x^2+y^2=25 and C_2:(x- )^2+y^2=16, where (5,9).

Question

Consider two circles ${C}_{1}:{x}^{2}+{y}^{2}=25$ and ${C}_{2}:(x-\alpha {)}^{2}+{y}^{2}=16$, where $\alpha \in (5,9)$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of ${C}_{1}$ and ${C}_{2}$ be ${\sin }^{-1}(\frac{\sqrt{63}}{8})$. If the length of common chord of ${C}_{1}$ and ${C}_{2}$ is $\beta $, then the value of $(\alpha \beta {)}^{2}$ equals \_\_\_\_\_\_\_.

Answer: 1575

Accepted Answer

At an intersection point $P$, $\cos(\angle O_1PO_2)=\sqrt{1-\tfrac{63}{64}}=\dfrac18$. By the law of cosines in $\triangle O_1PO_2$ (with $O_1P=5,\ O_2P=4$):

$$O_1O_2^2=25+16-2(5)(4)\left(\frac18\right)=41-5=36 \;\Rightarrow\; \alpha=6.$$

Radical axis: subtracting the circle equations gives $12x-45=0\Rightarrow x=\dfrac{15}{4}$. Half-chord:

$$\sqrt{25-\left(\frac{15}{4}\right)^2}=\sqrt{\frac{175}{16}}=\frac{5\sqrt7}{4} \;\Rightarrow\; \beta=\frac{5\sqrt7}{2}.$$

$$(\alpha\beta)^2=\left(6\times\frac{5\sqrt7}{2}\right)^2=(15\sqrt7)^2=1575.$$

Solutions · 1