2.1(1): Curve Equations and the Standard Equation of a Circle | Brief Solutions
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$(x+1)^2+(y-5)^2=17$. The center is the midpoint and $r^2=PQ^2/4$.
Set $A(-2,0),B(2,0)$. Then $2x^2+2y^2+8=16$, so $x^2+y^2=4$.
Put the origin at the chord midpoint: $y=\sqrt{841-x^2}-\sqrt{805}$, $-6\le x\le6$. The center is $(0,-\sqrt{805})\approx(0,-28.373)$.
$(x-3)^2+(y-4)^2=53$, since $r^2=CM^2$.
$(x+1)^2+(y-5)^2=17$; use the midpoint as the center.
$(1,-2)$;$\sqrt5$.
$x^2+y^2=9$.
$1:3$.
$(x+\tfrac32)^2+(y-3)^2=3$.
$(x-\sqrt2)^2+(y-1)^2=6$.
$(x-3)^2+(y-1)^2=5$ or $(x-3)^2+(y+1)^2=5$.
C
C
$(x+1)^2+(y+2)^2=10$. Intersect the perpendicular bisector $y=-2x-4$ with the given line to get $(-1,-2)$.
The center is $(0,-10.5)$ and the radius is $14.5$. At $x=-2$, the height is $\sqrt{14.5^2-2^2}-10.5\approx3.86$ m.
$(x-1)^2+(y-\tfrac13)^2=\tfrac{16}{9}$, using $P=(A+2B)/3$.
The locus is a circle of radius $\tfrac{15}{4}$, so its length is $\tfrac{15\pi}{2}$. Set $A(0,0),B(10,0)$ and complete the square.
A right triangle with hypotenuse $|ab|$: tangency gives $|ab|/\sqrt{a^2+b^2}=1$, hence $a^2+b^2=a^2b^2$.
$(x-1)^2+(y-1)^2=2$ or $(x+1)^2+(y+1)^2=2$. Solve $r^2=h^2+1=2k^2$ and $|h-2k|=1$ for the center $(h,k)$.