Circle practice questions — A-Level Maths

Centre $(h,k)=(2,-3)$ and $r=5$ (so $r^2=25$):

$(x-2{)}^2+(y-(-3){)}^2=25⟹(x-2{)}^2+(y+3{)}^2=25$.

We subtract the circle equations (the coefficient of $x^2$ and $y^2$ is $1$ in both, so they cancel): $$(x^2+y^2-4x-2y-4)-(x^2+y^2+2x-6y+1)=0⟹-6x+4y-5=0,$$ that is, multiplying by $-1$: $6x-4y+5=0$.

We check that the circles indeed intersect: solving the system, the intersection points have $x={\displaystyle \frac{1}{2}}\pm {\displaystyle \frac{\sqrt{299}}{13}}$ (real), so the common chord actually exists, and both points satisfy $6x-4y+5=0$. ✓

We bring the circle to canonical form: $x^2-6x+y^2+4y=23\Rightarrow (x-3{)}^2+(y+2{)}^2=23+9+4=36$, so the center is $C(3,-2)$ and the radius $r=6$.

Distance from $P(11,4)$ to the center: $d=\sqrt{(11-3{)}^2+(4+2{)}^2}=\sqrt{64+36}=\sqrt{100}=10$.

The length of the tangent is $\sqrt{d^2-r^2}=\sqrt{100-36}=\sqrt{64}=8$.

Check: the right triangle $P$–tangent point–center has legs $6$ (radius) and $8$ (tangent) and hypotenuse $10=d$, satisfying Pythagoras: $6^2+8^2=36+64=100={10}^2$. ✓

$AB$ is a horizontal segment ($y=2$) and $AC$ is a vertical segment ($x=1$), so the angle $\widehat{BAC}=90\mathrm{°}$. Triangle $ABC$ is right-angled at $A$, and the hypotenuse $BC$ is the diameter of the circumscribed circle (converse of Thales' theorem).

$$BC=\sqrt{(7-1{)}^2+(2-10{)}^2}=\sqrt{36+64}=\sqrt{100}=10⟹r=\frac{BC}{2}=5.$$

Check: the center is the midpoint of $BC$, $O(4,6)$, and $AO=\sqrt{(4-1{)}^2+(6-2{)}^2}=\sqrt{9+16}=5=r$. ✓

Let $M(x,y)$. The condition $MA={\displaystyle \frac{1}{2}}MB$ becomes, after squaring, $4M{A}^2=M{B}^2$: $$4(x^2+y^2)=(x-6{)}^2+y^2⟹4x^2+4y^2=x^2-12x+36+y^2$$ $$⟹3x^2+3y^2+12x-36=0⟹x^2+y^2+4x-12=0⟹(x+2{)}^2+y^2=16.$$

Unlike the case $MA=MB$ (which would give the perpendicular bisector, a line), a ratio $\mathrm{\ne }1$ produces a circle — the Apollonius circle, with center $(-2,0)$ and radius $\sqrt{16}=4$.

Check: the point $M(2,0)$ satisfies $(2+2{)}^2+0=16$ ✓, and $MA=2$, $MB=4$, so $MA=\frac{1}{2}MB$ ✓.