Circle practice questions — A-Level Maths

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

$(x - 2)^{2} + (y - (-3))^{2} = 25 \;\Longrightarrow\; (x - 2)^{2} + (y + 3)^{2} = 25$.

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

We check that the circles indeed intersect: solving the system, the intersection points have $x=\dfrac12\pm\dfrac{\sqrt{299}}{13}$ (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 $\widehat{BAC}=90°$. Triangle $ABC$ is right-angled at $A$, and the hypotenuse $BC$ is the diameter of the circumscribed circle (converse of Thales' theorem).

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

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 $MA=\dfrac12 MB$ becomes, after squaring, $4\,MA^2=MB^2$: $$4(x^2+y^2) = (x-6)^2+y^2 \implies 4x^2+4y^2 = x^2-12x+36+y^2$$ $$\implies 3x^2+3y^2+12x-36=0 \implies x^2+y^2+4x-12=0 \implies (x+2)^2+y^2=16.$$

Unlike the case $MA=MB$ (which would give the perpendicular bisector, a line), a ratio $\ne 1$ 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 $MA=\frac12 MB$ ✓.