Equation of a Circle

1. $(x-2)^2+(y+5)^2=11$.
2. $(x-1)^2+(y-2)^2=13$ using Pythagoras to find the radius.
3. Consider the radius joining the centre at (4,1) with the point (3,4). The gradient of this radius is -3 and so the tangent will have gradient $\frac{1}{3}$. The equation of the tangent so far is $y=\frac{1}{3}x+c$ and we find $c$ by inserting the coordinates (3,4) giving $c=3$. The equation of the tangent is $y=-\frac{1}{3}x+3$.

Answer:

1. Rearrange the equation by completing the square: $(x+3)^2-9+(y-2)^2-4+9=0$ or $(x+3)^2+(y-2)^2=4.$ Hence, the centre is at (-3,2) and the radius is 2.
2. Substituting $x=-4$ into this equation gives $1+(y-2)^2=4$. It follows that the y coordinates are given by $2\pm\sqrt{3}$ and so the coordinates are $(-4,2+\sqrt{3})$ and $(-4,2-\sqrt{3})$.

Always sketch when answering coordinate geometry questions. Since the line joining $(1,-2)$ and $(4,5)$ is a diameter, the line joining $(1,-2)$ and $(5,x)$ is at right angles to the line joining $(4,5)$ and $(5,x)$. The gradient of the line joining $(1,-2)$ and $(5,x)$ is $(x+2)/4$ (using rise over run). Hence, the line joining $(4,5)$ and $(5,x)$ is $-4/(x+2)$. Using rise over run to get another expression for this gradient we have $\frac{x-5}{1}=-\frac{4}{x+2}$ or $(x-5)(x+2)=-4$. It follows that solving the quadratic $x^2-3x-6=0$ gives $x=\frac{3\pm\sqrt{33}}{2}$. The y-coordinate is negative and so the exact value of x is $x=\frac{3-\sqrt{33}}{2}$.

Always begin by sketching the information given. We can find the point of intersection of the chord with the diameter since, if the intersection is at right angles, then they will intersect at the midpoint of the chord. It follows that the midpoint can be expressed as $\left(\frac{x-2}{2},-3\right)$. The line between this midpoint and the centre is perpendicular to the line between the midpoint and $(-2,-2)$, for instance. Firstly, the gradient of the line between the midpoint and the centre is $\frac{-\frac{5}{2}+3}{1-\frac{x-2}{2}}=\frac{0.5}{2-\frac{x}{2}}=\frac{1}{4-x}$. Secondly, the gradient between the midpoint and the point is $(-2,-2)$ is $\frac{-3+2}{\frac{x-2}{2}+2}=\frac{-2}{x+2}$. Note that the product of perpendicular gradients is -1 – this can be seen from $m\times -\frac{1}{m}$. It follows that $\frac{1}{4-x}\times \frac{-2}{x+2}=-1$ or $(4-x)(x+2)=2$. Solving this quadratic gives $x=1\pm\sqrt{7}$. From our sketch we can see that x is positive and so $x=1+\sqrt{7}$.