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 \(y=\dfrac{1}{2}x\) and this line intersects the vertical line at \(K\left(k, \dfrac{1}{2k}\right)\). Let \(L=(k,0)\) be the \(x\)-intercept of the vertical line. The area of triangle \(OKL\) must be \(\dfrac{18}{2} =9\). Therefore, \(\dfrac{1}{4}k^2=9\) 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.)
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, \[2\sqrt{2}=\dfrac{|c|}{\sqrt2}\] Therefore, we have \(c=\pm4\).
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 \[\begin{align*} 4c^2-4(2)(c^2-8)&=0\\ 4c^2-8c^2-64&=0 \\ 4c^2&=64\\ c^2&=16 \\ c&=\pm 4\end{align*}\]
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 \(20\).
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 \(\triangle AOB\) 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 \(-11\).
Since the slopes of \(AB\) and \(AC\) are \(1\) and \(-1\) respectively, the required line is vertical and its equation is \(x=0\).
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 \[\begin{align*} (x-3)^2+(mx-4)^2&=4\\ x^2-6x+9+m^2x^2-8mx+16&=4\\ (1+m^2)x^2-(6+8m)x+21&=0\end{align*}\] 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 \[\begin{align*} (6+8m)^2-4\cdot21\cdot(1+m^2)&=0\\ 36+96m+64m^{2}-84-84m^{2}&=0\\ -20m^{2}+96m-48&=0\\ m&=\dfrac{12\pm2\sqrt{21}}{5}\end{align*}\]
Therefore, both tangents we are considering have length \(\sqrt{21}\) and their slopes are \(\dfrac{12\pm2\sqrt{21}}{5}\).
The required set of points is the line that is the perpendicular bisector of the line segment \(CD\). Since \(CD\) has slope \(-\dfrac{1}{2}\) and midpoint \(M=\bigg(3\),\(\dfrac{3}{2}\bigg)\), the required line passes through \(M\) and has slope \(2\). The equation of the resulting line is \(4x-2y-9=0\).
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 \(M\bigg(\dfrac{x}{2},\dfrac{y}{2}\bigg)\) and \(N\bigg(\dfrac{a+d}{2},\dfrac{b}{2}\bigg)\). Now we are given that \(2MN=AW+DK\). Therefore, \[2\sqrt{\bigg(\dfrac{a+d-x}{2}\bigg)^2+\bigg(\dfrac{b-y}{2}\bigg)^2}=\sqrt{{(a-x)^2}+{(b-y)^2}}+d\] Squaring both sides and simplifying (using the fact that \(d \neq 0\)) gives \[\begin{align*} (a+d-x)^2+(b-y)^2&=(a-x)^2+(b-y)^2+2d\sqrt{(a-x)^2+(b-y)^2}+d^2 \\ 2d(a-x)&=2d\sqrt{(a-x)^2+(b-y)^2} \\ (a-x)&=\sqrt{(a-x)^2+(b-y)^2}\end{align*}\] Squaring both sides again and simplifying gives \[\begin{align*} (a-x)^{2} &= (a-x)^{2}+(b-y)^{2} \\ (b-y)^2&=0\end{align*}\]
This result gives \(b=y\) and implies that the slope of \(AW\) is \(0\). Therefore, \(AW\) is parallel to \(KD\).
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 \(\dfrac{a+b}{2}=4\) and \(\dfrac{c+d}{2}=2\). 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.
