Coordinate Geometry: Straight Lines and Circles — Practice

Question 1

Find the distance between the parallel lines $3x-4y+7=0$ and $3x-4y-8=0$.

Accepted Answer

On the first line, $(-1,1)$ works since $-3-4+7=0$.

Distance $=\dfrac{\left|3(-1)-4(1)-8ight|}{5}=\dfrac{15}{5}=3$.

Question 2

Find the bisector of the acute angle between $4x+3y-6=0$ and $5x+12y+9=0$.

Accepted Answer

Rewrite as $-4x-3y+6=0$ and $5x+12y+9=0$.

$a_1a_2+b_1b_2=-20-36=-56<0$, so the $+$ sign gives the acute bisector.

$\dfrac{-4x-3y+6}{5}=\dfrac{5x+12y+9}{13}$, giving $-52x-39y+78=25x+60y+45$, that is $77x+99y-33=0$, or $7x+9y-3=0$.

Question 3

Find the line through the intersection of $2x-3y+4=0$ and $3x+4y-5=0$ that passes through $(2,1)$.

Accepted Answer

$\left(2x-3y+4ight)+\lambda\left(3x+4y-5ight)=0$.

At $(2,1)$: $(4-3+4)+\lambda(6+4-5)=0$, so $5+5\lambda=0$ and $\lambda=-1$.

Subtracting: $-x-7y+9=0$, that is $x+7y=9$.

Question 4

A ray of light along $x+\sqrt3 y=\sqrt3$ strikes the $x$-axis and reflects. Find the equation of the reflected ray.

Accepted Answer

The incident ray meets the axis where $y=0$, at $\left(\sqrt3,0ight)$.

Reflecting the line in the $x$-axis replaces $y$ by $-y$: $x-\sqrt3y=\sqrt3$.

This passes through the same point, as it must.

Question 5

Find the orthocentre of the triangle with vertices $(0,0)$, $(4,0)$ and $(0,6)$.

Accepted Answer

The angle at the origin is a right angle, since the two legs lie along the axes.

For a right-angled triangle the orthocentre is the vertex containing the right angle.

So the orthocentre is $(0,0)$, and the circumcentre is the midpoint of the hypotenuse, $(2,3)$.

Question 6

A triangle has orthocentre $(3,5)$ and circumcentre $(6,2)$. Find its centroid.

Accepted Answer

From $H=3G-2O$ we get $G=\dfrac{H+2O}{3}$.

$G=\dfrac{(3,5)+2(6,2)}{3}=\dfrac{(15,9)}{3}=(5,3)$.

Question 7

Find the locus of a point whose distance from $(2,0)$ is twice its distance from the line $x=8$.

Accepted Answer

$(x-2)^{2}+y^{2}=4(x-8)^{2}$.

$x^{2}-4x+4+y^{2}=4x^{2}-64x+256$.

$3x^{2}-60x-y^{2}+252=0$, that is $3(x-10)^{2}-y^{2}=48$: a hyperbola centred at $(10,0)$.

Question 8

Find the equation of the circle with centre $(2,-3)$ that touches the line $3x-4y+1=0$.

Accepted Answer

Distance $=\dfrac{\left|6+12+1ight|}{5}=\dfrac{19}{5}$.

$(x-2)^{2}+(y+3)^{2}=\dfrac{361}{25}$.

Question 9

Find the length of the tangent from $(5,1)$ to $x^{2}+y^{2}-6x+4y-3=0$, and state whether the point is outside the circle.

Accepted Answer

$S_1=25+1-30+4-3=-3$.

Since $S_1<0$, the point lies **inside** the circle and no tangent exists.

Checking: the centre is $(3,-2)$ with radius $4$, and the distance from $(5,1)$ is $\sqrt{4+9}=\sqrt{13}<4$.

Question 10

How many common tangents do $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6x+8=0$ have?

Accepted Answer

Centres $(0,0)$ and $(3,0)$; radii $2$ and $1$.

$d=3=r_1+r_2$, so the circles touch externally.

There are $3$ common tangents.

Question 11

Show that $x^{2}+y^{2}-2x-4y+1=0$ and $x^{2}+y^{2}+4x+2y-4=0$ cut orthogonally.

Accepted Answer

$g_1=-1$, $f_1=-2$, $c_1=1$; $g_2=2$, $f_2=1$, $c_2=-4$.

$2g_1g_2+2f_1f_2=2(-2)+2(-2)=-8$.

$c_1+c_2=1-4=-3$.

These are not equal, so the circles do **not** cut orthogonally. The check itself is the answer, and it is worth performing before assuming a stated property.

Question 12

Find the circle through the intersections of $x^{2}+y^{2}-4=0$ and $x^{2}+y^{2}-2x-4y+4=0$ whose centre lies on the line $y=x$.

Accepted Answer

$S_1+\lambda S_2=(1+\lambda)\left(x^{2}+y^{2}ight)-2\lambda x-4\lambda y+(4\lambda-4)=0$.

Dividing, the centre is $\left(\dfrac{\lambda}{1+\lambda},\dfrac{2\lambda}{1+\lambda}ight)$.

On $y=x$ this needs $\lambda=2\lambda$, so $\lambda=0$ and the circle is $x^{2}+y^{2}=4$, whose centre is the origin.

Practice — Coordinate Geometry: Straight Lines and Circles

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