2.1(5): Relative Positions of Two Circles | Brief Solutions

They intersect: the center distance is $\sqrt5$, between the radius difference $1$ and sum $3$.

$a\in[1,121]$. Points of the second circle have distances from the origin between $1$ and $11$.

Centers $(2,-3),(3,-2)$ have distance $\sqrt2$, equal to the difference of radii $2\sqrt2,\sqrt2$. The contact point is $(4,-1)$.

(1) They intersect; the common chord lies on $x-y+4=0$ and has length $5\sqrt2$. (2) $x^2+y^2+3x+3y-16=0$, obtained by averaging the two circle equations.

Internal tangency at $m=7/2$; intersection for $7/6<m<7/2$. Compare $\sqrt{m^2-4m+8}$ with $|m-1|$ and $m+1$.

$(x-\frac{24}{5})^2+(y+\frac{18}{5})^2=1$. Extend the radius through the contact point by $1$.

$\sqrt{10}$. Subtraction gives the chord line $3x-y+1=0$; apply the chord-length formula.

A

B

A

$m\in(-\frac{12}{5},-\frac25)\cup(0,2)$.

$a=-2$.

$2x-y+5=0$.

$3x-y-9=0$.

$3$.

$\sqrt{23}$. The radical axis is $x+y=1$, at distance $1/\sqrt2$ from $(1,1)$.

$6x-2y-5=0$. Both radii are $\sqrt5$; bisect the segment between centers $(1,-1),(1/10,-7/10)$ perpendicularly.

$a\le-\sqrt3$ or $a\ge\sqrt3$, from $AC=\sqrt{1+a^2}\ge2$.

Reading the scanned length as $MN=\sqrt5$: $11x\pm\sqrt{583}\,y+9=0$. The contact chord is $x-ay+a^2-4=0$; $MN^2=16-64/(1+a^2)$ gives $a^2=53/11$.

$4$.

(1) Center locus $(x+1)^2+2y+4=0$; (2) $(x+1)^2+(y+2)^2=5$. The radical axis passes through $(-1,-1)$, yielding $m^2+2m+2n+5=0$; minimize $r_M^2=n^2+1$ at $(m,n)=(-1,-2)$.