Weekend Paper 1: Lines and Circles | Brief Solutions

$(x-2)^2+(y+1)^2=25$.

$\frac{13}{10}$.

$\pi-\arctan\frac12$.

$a=-3,-5$.

$x+y=2$;$y=x$.

$4x+y-6=0$;$3x+2y-7=0$.

$(x-\tfrac32)^2+y^2=\tfrac14$.

$\theta=\arccos\frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$.

$a\in[0,10]$.

$a\in\mathbb R\setminus\{-2,-1,1\}$.

$m=-\frac43$.

$7$.

B

C

A

A

Secant for $|m|>2\sqrt2$, tangent for equality, disjoint for $|m|<2\sqrt2$. Compare $6/\sqrt{1+m^2}$ with radius $2$.

$l_2:4x-3y+3=0$ or $3x-4y-3=0$. It passes through the reflected point $(-3,-3)$; apply the point-to-line distance formula.

$N(7,0),Q(3,4)$. Use $N=2F-M$ and reflection along the line normal.

Different: the squared sum is minimized at $(\frac{19}{10},\frac65)$, the projection of the midpoint of $MR$; the distance sum at $(\frac74,\frac32)$, the intersection of $RQ$ with the line.

$(x-3)^2+(y-3)^2=10$. The perpendicular bisector is $y=x$, giving center $(3,3)$.

The reflected line is $7x-4y-9=0$, through the reflected source $(-1,-4)$ and center $(3,3)$.

$x-y+4=0$; the slope is $1$.

$d=\frac{2\sqrt2}{n}$; $l_i:x-y+4\pm\frac{4i}{n}=0$, with the minus sign required for $i=n$. The side of each other line is unspecified, so either sign is possible independently.

No. The intercepts give $k_{n+1}=k_n-1/k_n$, and equal signs require $k_n^2>1$. Thus $k_{n+1}^2<k_n^2-1$, which cannot remain above $1$ indefinitely.