If P is a point on the circle S, Q is a point on a straight line L, and a given line M is …

Question

If $P$ is a point on the circle $S$, $Q$ is a point on a straight line $L$, and a given line $M$ is the perpendicular bisector of $PQ$, then $13$ times the sum of the abscissas of all such points $P$ is \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Here $S: x^2+y^2=4$, $L: 5x+y+2=0$, and the perpendicular bisector is $M: x-y+1=0$.

Solution:

Let $P=(x_1,y_1)$ on $S$, $Q=(x_2,y_2)$ on $L$ with $5x_2+y_2+2=0$. Midpoint of $PQ$ on $M$: $x_1+x_2-y_1-y_2=-2$. Since $PQ\perp M$ (slope of $M$ is $1$): $y_2-y_1=-(x_2-x_1)$.

Combining these gives $x_2=y_1-1$ and $y_2=x_1+1$. Substituting into $L$: $5(y_1-1)+(x_1+1)+2=0 \Rightarrow x_1=2-5y_1$.

Substituting into $S$: $(2-5y_1)^2+y_1^2=4 \Rightarrow 26y_1^2-20y_1=0 \Rightarrow y_1=0 \text{ or } \frac{10}{13}$.

$$x_1=2 \text{ or } -\frac{24}{13} \;\Rightarrow\; \text{sum}=2-\frac{24}{13}=\frac{2}{13}.$$

$$13\times\text{sum}=13\times\frac{2}{13}=2.$$

Solutions · 1