2.1(2): General Equation of a Circle | Brief Solutions
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(1) No real points; (2) the circle $(x-1)^2+(y-2)^2=5$; (3) the point $(1,2)$; (4) a circle only for $|a|>2$. Complete the square and check whether the squared radius is positive.
$(x-2)^2+(y-\tfrac34)^2=\tfrac{25}{16}$. The center lies on $x=2$ and is equidistant from all three points.
$t=-1$; the circle is $x^2+y^2-2x-1=0$. Substitution and Vieta give $\overrightarrow{OA}\cdot\overrightarrow{OB}=5x_1x_2=t$.
$x^2+y^2-2x-12=0$. Substitute the two points and use $-D-E=2$ to find $D=-2,E=0,F=-12$.
$x^2+y^2-5x-y+4=0$. Substitute the three points into the general equation.
$(x+a/2)^2+(y+1)^2=a^2/4-1$. A circle for $|a|>2$, the point $(-a/2,-1)$ for $|a|=2$, and no real points for $|a|<2$.
$a=-1$.
$x=1$;$x-\sqrt3y+2=0$.
$60^\circ$.
Tangent at the origin.
B
B
$x^2+y^2-ax-by=0$, obtained by substituting the three points.
For $\lambda=-1$, the line $y=x$; otherwise a circle with center $(\frac{\lambda}{1+\lambda},\frac1{1+\lambda})$ and radius $\frac{\sqrt{\lambda^2+1}}{|1+\lambda|}$.
B
D
$a\ne0$. Since $r^2=8(1/a-1/2)^2+2$, the minimum occurs at $a=2$: $(x-1)^2+(y+1)^2=2$.
For $k=1$, $x=0$; otherwise $(x-\frac{5(k^2+1)}{k^2-1})^2+y^2=\frac{100k^2}{(k^2-1)^2}$. Square the distance equation and complete the square.
(1) $-\frac17<t<1$; (2) $(x-\frac{24}{7})^2+(y+\frac{13}{49})^2=\frac{16}{7}$; (3) $t\in(-\frac17,0)\cup(\frac34,1)$. The center is $(t+3,4t^2-1)$ and $r^2=-7t^2+6t+1$; the outside condition is $t^2+1>r^2$.