A point (2h+1,,3k+2) lies on the circle x^2+y^2=4, while the point (h,k) lies on a circle …

Question

A point $(2h+1,\,3k+2)$ lies on the circle $x^2+y^2=4$, while the point $(h,k)$ lies on a circle $h^2+k^2=1$. Eliminating $h,k$ gives the locus of $(x,y)=(2h+1,3k+2)$:

$$\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=1$$

which is an ellipse with $a^2=9,\ b^2=4$, so $e^2 = 1-\dfrac{4}{9}=\dfrac{5}{9}$.

Answer: 9

Accepted Answer

Writing $x=2h+1,\ y=3k+2$, i.e. $h=\dfrac{x-1}{2},\ k=\dfrac{y-2}{3}$, and substituting into $h^2+k^2=1$ gives

$$\left(\frac{x-1}{2}\right)^2+\left(\frac{y-2}{3}\right)^2=1 \;\Rightarrow\; \frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=1.$$

This is an ellipse (shifted to centre $(1,2)$) with denominators $4$ and $9$; the larger denominator (semi-major axis squared) is $9$, which is the required value.

*(The exact phrasing of the blank quantity was corrupted in the source; the reconstructed value consistent with the stated answer is $9$, the semi-major-axis-squared of the derived ellipse.)*

Solutions · 1