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Let $S_1 : ( x+3)^2 + y^2 = 9$ and $S_2: ( x-5)^2 + y^2 = 16$ with centres $C_1$ and $C_2$ respectively. From a point $A$ of $S_2$ which is nearest to $C_1$, a variable chord is drawn to $S_1$.

1) Find the locus of mid point of the chord

2) Locus found in 1) cuts $S_1$ at $B$ and $C$, then line segment $BC$ subtends an angle on the major arc of circle $S_1$ is?

My work: Nearest point to $C_1$ on circle $S_2$ will be $(1,0)$ as both circles have centre on $x$-axis. Then I assumed that the midpoint of the variable chord to be $(h,k)$ and the end point of this chord to be $(x,y)$ which will be on circle. Then found $x$ and $y$ in terms of $h$ and $k$ and substituted in $S_1$ as $(x,y)$ lies on $S_1$. This will give locus of the midpoint. This will come to be: $x^2 + y^2 + 2x - 5/4 = 0$.

Now I solved this and $S_1$ to get $B$ and $C$. Then I calculated length of chord, distance of any point of circumference of $S_1$ and found the angle. But didn't got answer.

I think that the midpoint about which question is asking do not have end points $A$ and $(x,y)$ which i have assumed to find locus. My major problem is finding locus. If this is the problem then I want to know how to find required locus?

Robert Z
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Rohit Sinha
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2 Answers2

1

Let $P_1$ and $P_2$ be the intersection points of the line through $A:=(1,0)$ and the circle $S_1$ and let $M=(P_1+P_2)/2$ be the midpoint. Then the segments $C_1M$ and $M_1A$ are orthogonal (because the axis of the segment $P_1P_2$ goes through the center $C_1$). Hence $\triangle C_1MA$ is a right triangle and the locus you are looking for is a circle with diameter $C_1A$. Therefore its equation is $$(x+1)^2+y^2=2^2$$ (which is different from your $x^2 + y^2 + 2x - 5/4 = 0$ that is $(x+1)^2+y^2=\left(\frac{3}{2}\right)^2)$.

It follows that the intersection $B$ is $(-3/4,3/4\sqrt{7})$ and the subtended angle by the segment $BC$ is twice $\arccos(3/4)$.

Robert Z
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1

Intersections between the locus and $S_1$ are $A$ and $B$ which can easily be found solving the system of their equations $$A\left(-\frac{3}{4};\;-\frac{1}{4} \left(3 \sqrt{7}\right)\right),\quad B\left(-\frac{3}{4};\;\frac{1}{4} \left(3 \sqrt{7}\right)\right)$$ so $AB=\frac{3 \sqrt{7}}{2}$

Using sine theorem we get $$\frac{AB}{\sin\alpha}=2r$$ where $r$ is the radius of $S_1$, so $r=3$ and $$\sin\alpha=\frac{\sqrt{7}}{4};\quad \alpha\approx 41^{\circ}\,24'$$

Raffaele
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