Circles: past paper questions & worked solutions

$x^2+y^2+8x-14y-79=0$ becomes $(x+4)^2-16+(y-7)^2-49-79=0$.

$(x+4)^2+(y-7)^2=144$.

The centre is $(-4,7)$ and the radius is $12$.

The distance from $(-4,7)$ to $O(0,0)$ is $\sqrt{(-4)^2+7^2}=\sqrt{65}$.

Since $\sqrt{65}<12$, the origin is inside the circle.

The nearest point on the circle lies along the line from the centre to the origin, so $OP=12-\sqrt{65}$.

$x^2-6x=(x-3)^2-9$ and $y^2+5y=\left(y+\dfrac52\right)^2-\dfrac{25}{4}$.

The equation becomes $(x-3)^2+\left(y+\dfrac52\right)^2=\dfrac{225}{4}$.

The centre is $\left(3,-\dfrac52\right)$ and the radius is $\sqrt{\dfrac{225}{4}}=\dfrac{15}{2}$.

The centre of $C_2$ is $(-k,0)$ and its radius is $5$. Since the circles touch externally, the distance between centres is $\dfrac{15}{2}+5=\dfrac{25}{2}$.

$(k+3)^2+\left(\dfrac52\right)^2=\left(\dfrac{25}{2}\right)^2$, so $(k+3)^2=150$. Since $k>0$, $k+3=5\sqrt6$, hence $k=5\sqrt6-3$.

$x^2+y^2+8x-10y=29$ becomes $(x+4)^2-16+(y-5)^2-25=29$.

$(x+4)^2+(y-5)^2=70$.

The centre of $C_1$ is $(-4,5)$ and its radius is $\sqrt{70}$.

$C_2$ has equation $(x-5)^2+(y+8)^2=52$, so its centre is $(5,-8)$ and its radius is $\sqrt{52}=2\sqrt{13}$.

The distance between $(-4,5)$ and $(5,-8)$ is $\sqrt{9^2+(-13)^2}=\sqrt{250}=5\sqrt{10}$.

$\sqrt{70}+2\sqrt{13}=15.58\ldots$, while $5\sqrt{10}=15.81\ldots$.

The distance between centres is greater than the sum of the radii, so the circles are separate. Therefore they neither touch nor intersect.

The centre is $A(3,5)$ and the circle passes through $B(8,-7)$, so $r^2=(8-3)^2+(-7-5)^2=25+144=169$.

Hence $C$ has equation $(x-3)^2+(y-5)^2=169$.

Since $MN$ is parallel to the $x$-axis, its equation has the form $y=c$.

The chord length is $4\sqrt{22}$, so the half-length is $2\sqrt{22}$.

At height $y=c$, the horizontal half-length satisfies $169-(c-5)^2=(2\sqrt{22})^2=88$.

$(c-5)^2=81$, so $c=14$ or $c=-4$.

The chord lies above the $x$-axis, so $c=14$. Therefore the line through $M$ and $N$ is $y=14$.

The centre is the midpoint of $(-2,5)$ and $(4,15)$, so it is $\left(\dfrac{-2+4}{2},\dfrac{5+15}{2}\right)=(1,10)$.

Using $(-2,5)$, $r^2=(-2-1)^2+(5-10)^2=9+25=34$.

Therefore $C$ has equation $(x-1)^2+(y-10)^2=34$.

The nearest point to the $x$-axis is vertically below the centre by one radius.

The radius is $\sqrt{34}$, so the point is $\left(1,10-\sqrt{34}\right)$.

The centre is $(2,5)$ and $P=(8,-3)$. The radius is $\sqrt{(8-2)^2+(-3-5)^2}$.

$r=\sqrt{6^2+(-8)^2}=\sqrt{100}=10$.

With centre $(2,5)$ and radius $10$, the circle is $(x-2)^2+(y-5)^2=100$.

The gradient of the radius from the centre to $P$ is $\dfrac{-3-5}{8-2}=\dfrac{-8}{6}=-\dfrac43$.

The tangent is perpendicular to the radius, so its gradient is $\dfrac34$.

The tangent through $(8,-3)$ is $y+3=\dfrac34(x-8)$.

Multiplying by $4$ gives $4y+12=3x-24$, hence $3x-4y-36=0$.

Take $n=5$.

$(5+1)^3-5^3=216-125=91$.

$91=7\times13$, so the statement is false.

For $A(1,0)$ and $C(7,-6)$, the gradient of $CA$ is $\dfrac{0-(-6)}{1-7}=-1$.

For $B(3,-10)$ and $C(7,-6)$, the gradient of $CB$ is $\dfrac{-10-(-6)}{3-7}=1$.

The product of the gradients is $-1$, so $CA$ is perpendicular to $CB$ and $\angle ACB=90^\circ$.

An angle of $90^\circ$ at the circumference subtends a diameter, so $AB$ is a diameter of the circle.

$x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$.

The circle becomes $(x-3)^2+(y+4)^2=25-k$.

The centre is $(3,-4)$. Since the $x$-axis is tangent, the radius is the distance from the centre to the $x$-axis, which is $4$.

Thus $25-k=4^2=16$, so $k=9$.

With $k=9$, the circle is $(x-3)^2+(y+4)^2=16$. On the $x$-axis, $y=0$, giving $x=3$.

On the $y$-axis, $x=0$, so $9+(y+4)^2=16$. Hence $y=-4\pm\sqrt7$.

The vertical base on the $y$-axis has length $2\sqrt7$, and its perpendicular distance to $(3,0)$ is $3$. The area is $\dfrac12(2\sqrt7)(3)=3\sqrt7$.

The tangent has gradient $\dfrac12$, so the radius $PN$ has gradient $-2$.

Line $PN$ passes through $N(4,-1)$, so $y+1=-2(x-4)$. Therefore $y=-2x+7$.

At $P$, the tangent $y=\dfrac12x$ meets the radius $y=-2x+7$. Thus $\dfrac12x=-2x+7$, so $x=\dfrac{14}{5}$ and $y=\dfrac75$.

$r^2=\left(\dfrac{14}{5}-4\right)^2+\left(\dfrac75+1\right)^2=\left(-\dfrac65\right)^2+\left(\dfrac{12}{5}\right)^2=\dfrac{36}{5}$.

With centre $(4,-1)$ and radius squared $\dfrac{36}{5}$, the circle is $(x-4)^2+(y+1)^2=\dfrac{36}{5}$.

For $A(-1,2)$ and $B(11,6)$, the midpoint is $M\left(\dfrac{-1+11}{2},\dfrac{2+6}{2}\right)=(5,4)$.

The gradient of $AB$ is $\dfrac{6-2}{11-(-1)}=\dfrac{4}{12}=\dfrac13$.

The centre lies on the perpendicular bisector of the chord $AB$, so line $l$ has gradient $-3$ and passes through $M(5,4)$.

$y-4=-3(x-5)$, so $y=-3x+19$.

Since $P=(7,k)$ lies on $l$, $k=-3(7)+19=-2$.

The centre is $(7,-2)$. Using point $A$, $r^2=(7-(-1))^2+(-2-2)^2=8^2+(-4)^2=80$.

Therefore the circle is $(x-7)^2+(y+2)^2=80$.

The tangent $l$ has gradient $\dfrac52$, so the radius $XN$ is perpendicular to it and has gradient $-\dfrac25$.

The line through $X(4,-3)$ with gradient $-\dfrac25$ is $y+3=-\dfrac25(x-4)$.

Multiplying by $5$ gives $5y+15=-2x+8$, so $2x+5y+7=0$.

Use $y=\dfrac52x-\dfrac{55}{2}$ in $2x+5y+7=0$.

$2x+5\left(\dfrac52x-\dfrac{55}{2}\right)+7=0$. Multiplying by $2$ gives $4x+25x-275+14=0$, so $29x=261$ and $x=9$.

$y=\dfrac52(9)-\dfrac{55}{2}=\dfrac{45-55}{2}=-5$, so $N=(9,-5)$.

$XN^2=(9-4)^2+(-5+3)^2=5^2+(-2)^2=29$.

With centre $(4,-3)$ and radius squared $29$, the circle is $(x-4)^2+(y+3)^2=29$.

The gradient of $QP$ is $\dfrac{14-(-30)}{23-15}=\dfrac{44}{8}=\dfrac{11}{2}$.

The gradient of $QR$ is $\dfrac{-26-(-30)}{-7-15}=\dfrac4{-22}=-\dfrac2{11}$.

The product of the two gradients is $-1$, so $QP$ is perpendicular to $QR$ and $\angle PQR=90^\circ$.

Since the angle at $Q$ is a right angle, $PR$ is a diameter of the circle.

The centre is the midpoint of $P(23,14)$ and $R(-7,-26)$, so it is $(8,-6)$.

$PR=\sqrt{30^2+40^2}=50$, so the radius is $25$.

The greatest distance from $Q$ occurs at the point opposite $Q$ on the diameter through the centre. Thus $S=2(8,-6)-(15,-30)=(1,18)$.

The radius from the centre to $S$ has gradient $\dfrac{18-(-6)}{1-8}=-\dfrac{24}{7}$, so the tangent gradient is $\dfrac7{24}$.

$y-18=\dfrac7{24}(x-1)$, which rearranges to $7x-24y+425=0$.

$x^2+10x=(x+5)^2-25$ and $y^2-12y=(y-6)^2-36$.

The circle becomes $(x+5)^2+(y-6)^2=k+61$.

The centre is $(-5,6)$.

For a circle with positive radius, $k+61>0$, so $k>-61$.

If $QR$ is a diameter and $P$ lies on the circle, then $\angle QPR=90^\circ$.

So $(Q-P)\cdot(R-P)=0$. With $P=(p,0)$, $Q=(-2,10)$, and $R=(8,-14)$, this gives $(-2-p,10)\cdot(8-p,-14)=0$.

$(-2-p)(8-p)-140=0$, so $p^2-6p-156=0$.

$p=\dfrac{6\pm\sqrt{660}}{2}=3\pm\sqrt{165}$. Since $p$ is positive, $p=3+\sqrt{165}$.

$x^2+6x=(x+3)^2-9$ and $y^2-4y=(y-2)^2-4$.

The circle becomes $(x+3)^2+(y-2)^2=27$. Hence the centre is $(-3,2)$ and the radius is $\sqrt{27}=3\sqrt3$.

A horizontal tangent has equation $y=k$. The possible tangent levels are one radius above and below the centre, so $k=2\pm3\sqrt3$.

The chord has length $4$, so half the chord is $2$. If the chord is $y=p$, its perpendicular distance from the centre is $|p-2|$.

$(p-2)^2+2^2=(3\sqrt3)^2=27$, so $(p-2)^2=23$.

Since $p$ is negative, $p=2-\sqrt{23}$.

Since $AB$ is a diameter, the centre is the midpoint of $A(-2,18)$ and $B(14,6)$. This is $(6,12)$.

Using $A(-2,18)$, $r^2=(-2-6)^2+(18-12)^2=64+36=100$.

Therefore $C_1$ has equation $(x-6)^2+(y-12)^2=100$.

The distance from the origin to $(6,12)$ is $\sqrt{6^2+12^2}=6\sqrt5$.

Since $C_1$ has radius $10$, a circle centred at the origin can touch it with radius $6\sqrt5-10$ or $6\sqrt5+10$.

Thus $C_2$ can be $x^2+y^2=(6\sqrt5-10)^2=280-120\sqrt5$ or $x^2+y^2=(6\sqrt5+10)^2=280+120\sqrt5$.

$x^2+y^2+4x-30y+209=0$ becomes $(x+2)^2+(y-15)^2=20$.

The centre is $(-2,15)$ and the radius is $\sqrt{20}=2\sqrt5$.

The line $y=mx+1$ can be written as $mx-y+1=0$. Its distance from $(-2,15)$ must be $2\sqrt5$.

$\dfrac{|m(-2)-15+1|}{\sqrt{m^2+1}}=2\sqrt5$, so $\dfrac{|-2m-14|}{\sqrt{m^2+1}}=2\sqrt5$.

Squaring gives $4(m+7)^2=20(m^2+1)$. Hence $(m+7)^2=5(m^2+1)$, so $2m^2-7m-22=0$.

$2m^2-7m-22=0$ gives $m=-2$ or $m=\dfrac{11}{2}$.

A tangent is perpendicular to the radius at the point of contact.

For $m=-2$, the radius gradient is $\dfrac12$. The radius through the centre is $y-15=\dfrac12(x+2)$. Solving with $y=-2x+1$ gives $x=-6$ and $y=13$.

For $m=\dfrac{11}{2}$, the radius gradient is $-\dfrac2{11}$. The radius through the centre is $y-15=-\dfrac2{11}(x+2)$. Solving with $y=\dfrac{11}{2}x+1$ gives $x=\dfrac{12}{5}$ and $y=\dfrac{71}{5}$.

$x^2+y^2+8x-4y=0$ becomes $(x+4)^2-16+(y-2)^2-4=0$.

So $(x+4)^2+(y-2)^2=20$.

The centre is $(-4,2)$ and the radius is $\sqrt{20}=2\sqrt5$.

The tangent $x+2y+10=0$ has gradient $-\dfrac12$.

The radius to the point of contact is perpendicular to the tangent, so its gradient is $2$.

The radius line through the centre $(-4,2)$ is $y-2=2(x+4)$, so $y=2x+10$.

Substitute $y=2x+10$ into $x+2y+10=0$: $x+2(2x+10)+10=0$.

$5x+30=0$, so $x=-6$. Then $y=2(-6)+10=-2$, hence $P=(-6,-2)$.

The normal at $P$ is the radius line already found, so its equation is $y=2x+10$.

From $(x-k)^2+(y-2k)^2=k+7$, the centre is $(k,2k)$ and the radius is $\sqrt{k+7}$.

Since $P(2,3)$ lies on the circle, $(2-k)^2+(3-2k)^2=k+7$.

Expanding gives $k^2-4k+4+4k^2-12k+9=k+7$, hence $5k^2-17k+6=0$.

$5k^2-17k+6=(5k-2)(k-3)$, so $k=\dfrac25$ or $k=3$.

For the tangent calculation, $k<2$, so the centre is $\left(\dfrac25,\dfrac45\right)$.

The radius from the centre to $P$ has gradient $\dfrac{3-4/5}{2-2/5}=\dfrac{11}{8}$, so the tangent gradient is $-\dfrac8{11}$.

The tangent is $y-3=-\dfrac8{11}(x-2)$. Setting $y=0$ gives $x=\dfrac{49}{8}$, so $T=\left(\dfrac{49}{8},0\right)$.

Triangle $OPT$ has base $\dfrac{49}{8}$ on the $x$-axis and height $3$, so its area is $\dfrac12\cdot\dfrac{49}{8}\cdot3=\dfrac{147}{16}$.

For $y=x^3-5x^2+3x+14$, $\dfrac{dy}{dx}=3x^2-10x+3$.

$3x^2-10x+3=0$ factors as $(3x-1)(x-3)=0$, so $x=\dfrac13$ or $x=3$.

$\dfrac{d^2y}{dx^2}=6x-10$. At $x=3$, this is positive, so the minimum turning point has $x=3$.

Substitute $x=3$: $y=27-45+9+14=5$. Hence $T=(3,5)$.

At $x=2$ on $C_1$, $y=8-20+6+14=8$, so $A=(2,8)$.

The centre of $C_2$ is $T=(3,5)$ and it passes through $A=(2,8)$. Thus $r^2=(2-3)^2+(8-5)^2=10$.

Therefore $C_2$ has equation $(x-3)^2+(y-5)^2=10$.

The radius $TA$ has gradient $\dfrac{8-5}{2-3}=-3$, so the tangent at $A$ has gradient $\dfrac13$.

Using $A=(2,8)$, $y-8=\dfrac13(x-2)$, so $y=\dfrac13x+\dfrac{22}{3}$.

From $x=0$ to $x=2$, the cubic is above the line. The required area is $\displaystyle \int_0^2 \left[x^3-5x^2+3x+14-\left(\dfrac13x+\dfrac{22}{3}\right)\right]dx$.

The integrand is $x^3-5x^2+\dfrac83x+\dfrac{20}{3}$.

$\displaystyle \left[\dfrac{x^4}{4}-\dfrac{5x^3}{3}+\dfrac{4x^2}{3}+\dfrac{20x}{3}\right]_0^2=4-\dfrac{40}{3}+\dfrac{16}{3}+\dfrac{40}{3}=\dfrac{28}{3}$.

The circle has centre $(3,5)$ and radius $r$, so $(x-3)^2+(y-5)^2=r^2$.

On the line, $y=2x+k$. Substitution gives $(x-3)^2+(2x+k-5)^2=r^2$.

$x^2-6x+9+4x^2+4kx-20x+k^2-10k+25-r^2=0$.

Collecting terms gives $5x^2+(4k-26)x+k^2-10k+34-r^2=0$.

For a tangent, the quadratic has one repeated root, so its discriminant is zero.

$(4k-26)^2-20(k^2-10k+34-r^2)=0$.

This reduces to $20r^2-4k^2-8k-4=0$, hence $5r^2=(k+1)^2$. Therefore $p=1$.

For the repeated root, $x_B=-\dfrac{4k-26}{2\cdot5}=\dfrac{13-2k}{5}$.

The point $A$ is $(0,k)$ and $B$ lies on a line of gradient $2$, so $AB=\sqrt{x_B^2+(2x_B)^2}=\sqrt5\,x_B$.

From $5r^2=(k+1)^2$, $r=\dfrac{k+1}{\sqrt5}$ in the shown configuration.

$\sqrt5\left(\dfrac{13-2k}{5}\right)=2\left(\dfrac{k+1}{\sqrt5}\right)$.

Multiplying by $\sqrt5$ gives $13-2k=2k+2$, so $k=\dfrac{11}{4}$.