2.1(4): Lines Tangent to Circles | Brief Solutions

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Labels: E = worked example, C = class practice, H = homework; P1-P4 refer to original PDF pages. Other labels follow the original question or section numbering.

$(x+1)^2+(y-2)^2=13$; use the center-to-line distance as the radius.

$x=2$ or $x-\sqrt3y+4=0$. Set the line distance to $2$ and include the vertical case.

$x_0x+y_0y=r^2$. The radius vector $(x_0,y_0)$ is normal to the tangent.

$(x-3)^2+(y-1)^2=9$ or $(x+3)^2+(y+1)^2=9$. Set the center to $(3t,t)$ and use $7t^2=7$.

Two if $\frac{\sqrt5}{2}<|k|\le\frac32$; one if $|k|=\frac{\sqrt5}{2}$ or $|k|>\frac32$; none if $|k|<\frac{\sqrt5}{2}$. Use the upper-semicircle tangent and endpoint slopes.

Through $A$: $x+\sqrt3y=4$; through $B$: $x=2$ or $5x-12y+26=0$. Use the point-of-contact formula and the distance-to-line condition.

$(x-1)^2+(y-2)^2=25$, since $r=|4+6-35|/5=5$.

$-3x+4y=25$.

$x=5$;$y=-5$.

$x=2$;$3x-4y+10=0$.

$(x-\frac{24}{5})^2+(y+\frac{18}{5})^2=1$.

(1) $a^2+b^2=r^2$;(2) $b=0$;(3) $|b|=r$;(4) $|a|=|b|=r$.

$(x-5)^2+(y-4)^2=25$. The center is $(a,4)$ with $a^2=4^2+3^2$ and $a>0$, hence $a=5$.

$(x-1)^2+(y+2)^2=2$ or $(x-9)^2+(y+18)^2=338$. With center $(a,-2a)$, equating the two radius formulas gives $(a-1)(a-9)=0$.

$a\Rightarrow3$;$b\Rightarrow1$;$c\Rightarrow2$.

Incident lines: $4x+3y+3=0$ or $3x+4y-3=0$. Reflect the source to $(-3,-3)$; tangent slopes are $4/3,3/4$, then reflect back.

(1) $(x-1)(y-1)=\tfrac12$, $x>1,y>1$; (2) $S_{\min}=3+2\sqrt2$. Tangency with the center below the line gives $ab-2a-2b+2=0$ for intercepts $a,b$; apply AM-GM to $x-1,y-1$.

$1\le m<\sqrt2$. The boundaries come from an endpoint and tangency.

$\frac5{12}<k\le\frac34$. The tangent slope is $5/12$ and the line through the left endpoint has slope $3/4$.

For $a=0$: $(x-4)^2+(y-\tfrac{17}{2})^2=\tfrac{289}{4}$; for $a=1$: $(x-2)^2+(y-\tfrac52)^2=\tfrac{25}{4}$. With center $(h,(h^2+1)/2)$, solve $(1-a)h^2-8h+16+a^2-a=0$, checking both a repeated root and the linear degeneration.

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2.1(4): Lines Tangent to Circles | Brief Solutions

E1Circle centered at $(-1,2)$ tangent to $2x-3y-5=0$.

E2Tangents from $(2,2\sqrt3)$ to $x^2+y^2=4$.

E3Tangent to $x^2+y^2=r^2$ at $M(x_0,y_0)$.

E4Circle tangent to the $y$ axis, center on $x-3y=0$, chord length $2\sqrt7$ on $y=x$.

E5Number of solutions of $\sqrt{4-x^2}-kx-3=0$.

C1Tangents to $x^2+y^2=4$ through $A(1,\sqrt3)$ and $B(2,3)$.

C2Circle centered at $(1,2)$ tangent to $4x+3y-35=0$.

H1Tangent through $(-3,4)$ to $x^2+y^2=25$.

H2Tangents from $(5,-5)$ to $x^2+y^2=25$.

H3Tangents from $(2,4)$ to $x^2+y^2=4$.

H4Radius-1 circle externally tangent to $x^2+y^2=25$ at $(4,-3)$.

H5Conditions for $(x-a)^2+(y-b)^2=r^2$: through the origin, center on the $x$ axis, tangent to the $x$ axis, tangent to both axes.

H6Circle centered at $(2,-1)$ tangent to $3x-4y+5=0$.

H7Circle tangent to the $y$ axis at $(0,4)$ and cutting a length-6 chord on the positive $x$ axis.

H8Circle through $(2,-1)$, center on $2x+y=0$, tangent to $x-y-1=0$.

H9Match positions of $M$ with the relation of $x_0x+y_0y=r^2$ to the circle.

H10A ray from $(-3,3)$ reflects off the $x$ axis and is tangent to $(x-2)^2+(y-2)^2=1$.

H11Upper tangents to the radius-1 circle centered at $(1,1)$ meet the positive axes at $A,B$: midpoint locus and minimum area.

H12.1$y=x+m$ has two intersections with $y=\sqrt{1-x^2}$.

H12.2$y=1+\sqrt{4-x^2}$ and $y=k(x-2)+4$ have two intersections.

H13Exactly one circle passes through $A(0,1),B(4,a)$ and is tangent to the $x$ axis.