Analytic Geometry
Solutions

1. Using the line segment from $O(0,0)$ to $C(9,0)$ as the base and noting that the height is $4$, the area of triangle $OCD$ is $18$. We let the vertical line be $x=k$. The line from $O(0,0)$ to $D(8,4)$ is $y={\displaystyle \frac{1}{2}}x$ and this line intersects the vertical line at $K(k,{\displaystyle \frac{1}{2k}})$. Let $L=(k,0)$ be the $x$-intercept of the vertical line. The area of triangle $OKL$ must be ${\displaystyle \frac{18}{2}}=9$. Therefore, ${\displaystyle \frac{1}{4}}{k}^2=9$ and so the vertical line required is $x=6$. (The value of $k=-6$ is not admissible since the line $x=-6$ does not intersect the triangle.)

2. There are several ways to solve this question. We will give two solutions using analytic geometry.

   Solution 1

   If the line is tangent to the circle, then the distance from the centre $(0,0)$ to the line $y=x+c$ (or $x-y+c=0$) equals the radius of the circle which is $2\sqrt{2}$. Using the formula for distance from a point to a line, $$2\sqrt{2}={\displaystyle \frac{|c|}{\sqrt{2}}}$$ Therefore, we have $c=\pm 4$.

   Solution 2

   We substitute $y=x+c$ into the equation of the circle to obtain $x^2+(x+c{)}^2=8$. Expanding we obtain $2x^2+2xc+c^2-8=0$. If the line is tangent to the circle, then we need to find values of $c$ such that this quadratic has exactly one root. Therefore, it must be the case that $$\begin{array}{rl}4c^2-4(2)(c^2-8)& =0\\ 4c^2-8c^2-64& =0\\ 4c^2& =64\\ c^2& =16\\ c& =\pm 4\end{array}$$

   

3. There are two circles, the first with centre $(0,0)$ and radius $k$, and the second with centre $(5,-12)$ and radius $7$. The distance between the centres can be calculated to be $\sqrt{(-5{)}^2+(12{)}^2}=13$. Since the two circles intersect only once, they can be either externally or internally tangent.

   If they are externally tangent, $k+7=13$ and so $k=6$.

   

   If they are internally tangent, $|k-7|=13$ and so $k=20$ or $-6$. But the radius must be positive so $k=6$ or $20$.

   

4. Solution 1

   All lines that cut a circle in half pass through the centre. Now the perpendicular bisector of any chord passes through the centre. If we consider the vertical chord from $(0,0)$ to $(0,10)$, the perpendicular bisector is the horizontal line $y=5$. Similarly, if we consider the horizontal chord from $(0,0)$ to $(8,0)$, the perpendicular bisector is the vertical line $x=4$. These two lines intersect at $(4,5)$ and therefore, the centre is $(4,5)$. We require the $y$-intercept of the line through $(4,5)$ and $P(2,-3)$. This line is $y=4x-11$ and the $y$-intercept is $-11$.

   Solution 2

   Observe that $\mathrm{△}AOB$ is right-angled at $O$, thus $AB$ is the diameter of the circle, and its midpoint $(4,5)$ is the centre of the circle. As in solution 1, we require the $y$-intercept of the line that goes through $(4,5)$ and $P(2,-3)$. This line is $y=4x-11$ and the $y$-intercept is $-11$.

   Solution 3

   The general equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h{)}^2-(y-k{)}^2=r^2$. Substituting in the point $(0,0)$ we obtain the equation $h^2+k^2=r^2$. Substituting in the point $(0,10)$ we obtain the equation $h^2+(10-k{)}^2=r^2$. Therefore, $(10-k{)}^2=k^2$ which gives $k=5$.

   Substituting the point $(8,0)$ into the equation of the circle gives $(8-h{)}^2+k^2=r^2$. Therefore, $(8-h{)}^2=h^2$ which gives $h=4$. As in solution 1, we require the $y$-intercept of the line that goes through $(4,5)$ and $P(2,-3)$. This line is $y=4x-11$ and the $y$-intercept is $-11$.

   

5. Since the slopes of $AB$ and $AC$ are $1$ and $-1$ respectively, the required line is vertical and its equation is $x=0$.

   

6. Consider a triangle with the following vertices: the origin, the centre of the circle and a point of tangency of one of the two tangents we are considering. Since there are two tangents, we have two such triangles. A tangent is perpendicular to the radius at the point of tangency and so these two triangles are right-angled. The two known sides of one of these right triangles are the radius $2$ and the segment from $(0,0)$ to $(3,4)$ which has length $5$. Thus, the other side has length $\sqrt{21}$.

   

   Since each of the tangents pass through the origin, their equations are of the form $y=mx$. We are interested in values of $m$ for which the line $y=mx$ intersects the circle only once. Substituting into the equation of the circle we get $$\begin{array}{rl}(x-3{)}^2+(mx-4{)}^2& =4\\ x^2-6x+9+m^2{x}^2-8mx+16& =4\\ (1+m^2)x^2-(6+8m)x+21& =0\end{array}$$ Now this quadratic will have one solution when its discriminant is zero. Thus, we are looking for values of $m$ that give a discriminant of $0$. So $$\begin{array}{rl}(6+8m{)}^2-4\cdot 21\cdot (1+m^2)& =0\\ 36+96m+64m^2-84-84m^2& =0\\ -20m^2+96m-48& =0\\ m& ={\displaystyle \frac{12\pm 2\sqrt{21}}{5}}\end{array}$$

   Therefore, both tangents we are considering have length $\sqrt{21}$ and their slopes are ${\displaystyle \frac{12\pm 2\sqrt{21}}{5}}$.

7. The required set of points is the line that is the perpendicular bisector of the line segment $CD$. Since $CD$ has slope $-{\displaystyle \frac{1}{2}}$ and midpoint $M=(3$,${\displaystyle \frac{3}{2}})$, the required line passes through $M$ and has slope $2$. The equation of the resulting line is $4x-2y-9=0$.

8. We present the solution that uses analytic geometry most directly. Let the coordinates of the points be $K(0,0)$, $W(x,y)$, $A(a,b)$ and $D(d,0)$. Therefore, the coordinates of $M$ and $N$ are $M({\displaystyle \frac{x}{2}},{\displaystyle \frac{y}{2}})$ and $N({\displaystyle \frac{a+d}{2}},{\displaystyle \frac{b}{2}})$. Now we are given that $2MN=AW+DK$. Therefore, $$2\sqrt{({\displaystyle \frac{a+d-x}{2}}{)}^2+({\displaystyle \frac{b-y}{2}}{)}^2}=\sqrt{(a-x{)}^2+(b-y{)}^2}+d$$ Squaring both sides and simplifying (using the fact that $d\ne 0$) gives $$\begin{array}{rl}(a+d-x{)}^2+(b-y{)}^2& =(a-x{)}^2+(b-y{)}^2+2d\sqrt{(a-x{)}^2+(b-y{)}^2}+d^2\\ 2d(a-x)& =2d\sqrt{(a-x{)}^2+(b-y{)}^2}\\ (a-x)& =\sqrt{(a-x{)}^2+(b-y{)}^2}\end{array}$$ Squaring both sides again and simplifying gives $$\begin{array}{rl}(a-x{)}^2& =(a-x{)}^2+(b-y{)}^2\\ (b-y{)}^2& =0\end{array}$$

   This result gives $b=y$ and implies that the slope of $AW$ is $0$. Therefore, $AW$ is parallel to $KD$.

9. Let the coordinates of $A$ and $B$ be $(a,c)$ and $(b,d),$ respectively. Point $A$ satisfies the equation of the first line and point $B$ satisfies the equation of the second line and so $4a+3c-48=0$ and $b+3d+10=0$ . Moreover, since $(4,2)$ is the midpoint, we know ${\displaystyle \frac{a+b}{2}}=4$ and ${\displaystyle \frac{c+d}{2}}=2$. Thus, $b=8-a$ and $d=4-c$. Substituting these into the second equation above and simplifying we obtain $-a-3c+30=0$. Adding this equation to the first equation gives $3a-18=0$ and so $a=6$. From the first equation we obtain that $c=8$. Therefore, $b=2$ and $d=-4$. So the coordinates of $A$ and $B$ are $(6,8)$ and $B(2,-4)$, respectively.