Circle flash cards

1. 1.Write the general equation of a circle and identify its center and radius.

   The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$, with center $(-g,-f)$ and radius $r=\sqrt{g^2+f^2-c}$ (valid provided $g^2+f^2-c\ge 0$).

   Hint: Compare coefficients of $x$ and $y$ terms.

2. 2.What is the condition for a line $y=mx+c$ to be tangent to the circle $x^2+y^2=a^2$?

   The line is tangent when $c^2=a^2(1+m^2)$, and the equation of the tangent lines with slope $m$ is $y=mx\pm a\sqrt{1+m^2}$.

   Hint: Set perpendicular distance from center equal to radius.

3. 3.State the equation of the chord of contact of tangents drawn from an external point $(x_1,y_1)$ to circle $x^2+y^2=a^2$.

   The chord of contact is $xx_1+yy_1=a^2$, obtained by replacing $x^2\to xx_1$ and $y^2\to yy_1$ in the circle's equation.

   Hint: Same substitution rule as tangent equation $T=0$.

4. 4.How do you determine the relative position of two circles with radii $r_1,r_2$ and distance between centers $d$?

   Circles touch externally if $d=r_1+r_2$; touch internally if $d=|r_1-r_2|$; intersect at two points if $|r_1-r_2|one circle lies inside the other (no common point) if$d<|r_1-r_2|$; and the circles are separate (non-intersecting, one outside the other) if $d>r_1+r_2$.

   Hint: Compare $d$ with the sum and difference of radii.

5. 5.What is the equation of the family of circles through the intersection of circle $S=0$ and line $L=0$ (or circle $S_1=0$ and $S_2=0$)?

   $S+\lambda L=0$ represents the family of circles through the points of intersection of $S=0$ and $L=0$, for any real $\lambda$. Similarly, $S_1+\lambda S_2=0$ (with $\lambda\neq -1$) represents the family of circles through the common points of $S_1=0$ and $S_2=0$; the value $\lambda=-1$ is excluded since it gives the radical axis (a line, not a circle).

   Hint: Radical axis based family; $\lambda$ is fixed by an extra condition.

6. 6.Find the circle through the intersection of $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+2=0$ that also passes through the origin.

   The radical axis is $S_1-S_2=-4x+4=0\Rightarrow x=1$. Take the family $S_1+\lambda(x-1)=0$; substituting $(0,0)$ gives $6-\lambda=0\Rightarrow\lambda=6$. This yields $x^2+y^2+4x-6y=0$.

   Hint: Use the family $S_1+\lambda L=0$ where $L$ is the radical axis.

7. 7.One member of the family $x^2+y^2-6x+2y+4+\lambda(x^2+y^2+2x-4y-6)=0$ has its center on the line $x+y=0$. Find $\lambda$ and the center.

   Dividing by $(1+\lambda)$, the center is $\left(\dfrac{3-\lambda}{1+\lambda},\dfrac{2\lambda-1}{1+\lambda}\right)$. Setting the coordinate sum to zero: $\dfrac{(3-\lambda)+(2\lambda-1)}{1+\lambda}=0\Rightarrow\lambda+2=0\Rightarrow\lambda=-2$. The center is then $(-5,5)$.

   Hint: Write the center of the combined family in terms of $\lambda$ first.

8. 8.Show that every circle of the family $x^2+y^2-2x+\lambda y=0$ (parameter $\lambda$) passes through two fixed points, and find them.

   This is $S+\lambda L=0$ with $S:x^2+y^2-2x=0$ and $L:y=0$. The fixed points are the intersections of $S=0$ and $L=0$: setting $y=0$ gives $x^2-2x=0\Rightarrow x=0,2$. So every member passes through $(0,0)$ and $(2,0)$, independent of $\lambda$.

   Hint: Set $\lambda=0$ and consider what condition is preserved for all $\lambda$.

9. 9.Find the circle through the intersection of $x^2+y^2-x-y=0$ and $x^2+y^2-2x-4y+3=0$ that also passes through $(1,2)$.

   With $S_1,S_2$ as given, $S_1(1,2)=2$ and $S_2(1,2)=-2$. The family $S_1+\lambda S_2=0$ gives $2-2\lambda=0\Rightarrow\lambda=1$. Adding: $S_1+S_2=2x^2+2y^2-3x-5y+3=0$, i.e. $2x^2+2y^2-3x-5y+3=0$.

   Hint: Plug the given point into both circle expressions to solve for $\lambda$.

10. 10.For the family $S_1+\lambda S_2=0$ where $S_1:x^2+y^2-4x-6y-12=0$ and $S_2:x^2+y^2+6x+4y-12=0$, for what value of $\lambda$ does the family degenerate into a straight line?

    The coefficient of $x^2$ (and $y^2$) is $(1+\lambda)$, which vanishes only at $\lambda=-1$. At $\lambda=-1$ the family becomes $S_1-S_2=-10x-10y=0$, i.e. the radical axis $x+y=0$.

    Hint: Ask when the quadratic ($x^2,y^2$) terms cancel out.

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