Geometry: lines and circles in coordinates

Answer: $(-1, -2)$

Complete the square in $x$ and in $y\mathclose{:}$ $\left(x + 1\right)^{2} + \left(y + 2\right)^{2} = 1 + 4 - 1 = 4\mathclose{.}$ So the centre is $(-1, -2)$ and the radius is $\sqrt{4} = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Copied the signs straight from the equation → $(1, 2)$
• Did not halve the coefficients → $(-2, -4)$

Answer: $1$

The centre is $(-1, -1)$ and the radius $12\mathclose{.}$ The point is $\sqrt{(-5)^2 + (-12)^2} = 13$ from the centre, so it lies outside the circle, and the nearest point of the circle is on the line through the centre: $|13 - 12| = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the distance to the centre → $13$
• Added the radius instead of subtracting it → $25$

Answer: $-\dfrac{2}{5}$

Perpendicular slopes multiply to $-1\mathclose{:}$ turn $\dfrac{5}{2}$ over and change its sign, giving $-\dfrac{2}{5}\mathclose{.}$ (Check: $\dfrac{5}{2} \times \left(-\dfrac{2}{5}\right) = -1\mathclose{.})$

Common mistakes, and the answer each one gives:

• Turned the slope over but kept its sign → $\frac{2}{5}$
• Changed the sign but did not turn the slope over → $- \frac{5}{2}$

Answer: $30$

Each slope is the tangent of the line's angle with the $x$-axis: $150^\circ$ and $120^\circ\mathclose{.}$ The lines meet at the difference, $30^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the obtuse angle instead of the acute one → $150$

Answer: $13$

The horizontal gap is $|3 - (-2)| = 5$ and the vertical gap is $|4 - (-8)| = 12\mathclose{,}$ so Pythagoras gives $\sqrt{5^2 + 12^2} = \sqrt{169} = 13\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two gaps instead of using Pythagoras → $17$
• Forgot the square root → $169$

Answer: $\dfrac{31}{10}$

For $ax + by + c = 0$ the distance is $\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\mathclose{.}$ Here the top is $|6 \cdot 4 + (-8) \cdot 0 + 7| = 31$ and the bottom $\sqrt{100} = 10\mathclose{,}$ so the distance is $\dfrac{31}{10}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by $\sqrt{100}$ → $31$
• Divided by $100$ instead of by its square root → $\frac{31}{100}$

Answer: $6$

Complete the square in $x$ and in $y\mathclose{:}$ $\left(x + 4\right)^{2} + \left(y + 6\right)^{2} = 16 + 36 - 16 = 36\mathclose{.}$ So the centre is $(-4, -6)$ and the radius is $\sqrt{36} = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $r^2$ instead of $r$ → $36$

Answer: $y = \frac{8 x}{15} - \frac{59}{3}$

The radius to $P$ runs from the centre $(-3, -2)$ to $(5, -17)\mathclose{,}$ so its slope is $\dfrac{-15}{8} = -\dfrac{15}{8}\mathclose{.}$ The tangent is perpendicular to it, with slope $\dfrac{8}{15}\mathclose{.}$ Through $P\mathclose{:}$ $y = \dfrac{8}{15}\left(x - 5\right) - 17\mathclose{,}$ which is $y = \frac{8 x}{15} - \frac{59}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the slope of the radius itself → $y = - \frac{15 x}{8} - \frac{61}{8}$
• Turned the radius's slope over but kept its sign → $y = - \frac{8 x}{15} - \frac{43}{3}$

Answer: $(0, -6)$

Substitute the line into the circle: a quadratic in $x\mathclose{,}$ whose roots are $x = -1$ and $x = 0\mathclose{.}$ The line then gives $y\mathclose{:}$ the points are $(-1, -7)$ and $(0, -6)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other intersection point → $(-1, -7)$
• Swapped the coordinates → $(-6, 0)$

Answer: $y = x + 1$

The given line has slope $-1\mathclose{,}$ so a perpendicular one has slope $1\mathclose{.}$ Through $(-3, -2)\mathclose{:}$ $y = \left(x + 3\right) - 2\mathclose{,}$ which is $y = x + 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the same slope, which gives a PARALLEL line → $y = - x - 5$

Answer: $120$

The slope is the tangent of that angle: $\tan\theta = - \sqrt{3}\mathclose{,}$ so $\theta = 120^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the angle in radians → $\frac{2\pi}{3}$
• Gave the acute angle with the axis → $60$

Answer: $\left(x - 1\right)^{2} + \left(y + 4\right)^{2} = 169$

The radius is the distance from the centre to the point: $\sqrt{(-12)^2 + 5^2} = \sqrt{169} = 13\mathclose{.}$ So the circle is $\left(x - 1\right)^{2} + \left(y + 4\right)^{2} = 169\mathclose{.}$

Common mistakes, and the answer each one gives:

• Put $r$ instead of $r^2$ on the right → $\left(x - 1\right)^{2} + \left(y + 4\right)^{2} = 13$
• Put the centre in with the wrong signs → $\left(x + 1\right)^{2} + \left(y - 4\right)^{2} = 169$

Answer: $(4, 5)$

Subtract one equation from the other: the squares cancel and a straight line is left — the common chord, $x = 4\mathclose{.}$ Put that into either circle: the points are $(4, 5)$ and $(4, -3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other intersection point → $(4, -3)$
• Swapped the coordinates → $(5, 4)$

Answer: $17$

The horizontal gap is $|-11 - (-3)| = 8$ and the vertical gap is $|-10 - 5| = 15\mathclose{,}$ so Pythagoras gives $\sqrt{8^2 + 15^2} = \sqrt{289} = 17\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two gaps instead of using Pythagoras → $23$
• Forgot the square root → $289$

Answer: $-2$

Perpendicular slopes multiply to $-1\mathclose{:}$ turn $\dfrac{1}{2}$ over and change its sign, giving $-2\mathclose{.}$ (Check: $\dfrac{1}{2} \times \left(-2\right) = -1\mathclose{.})$

Common mistakes, and the answer each one gives:

• Turned the slope over but kept its sign → $2$
• Changed the sign but did not turn the slope over → $- \frac{1}{2}$

Answer: $60$

Each slope is the tangent of the line's angle with the $x$-axis: $60^\circ$ and $0^\circ\mathclose{.}$ The lines meet at the difference, $60^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the obtuse angle instead of the acute one → $120$

Answer: $\dfrac{81}{13}$

For $ax + by + c = 0$ the distance is $\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\mathclose{.}$ Here the top is $|5 \cdot 6 + (-12) \cdot (-5) - 9| = 81$ and the bottom $\sqrt{169} = 13\mathclose{,}$ so the distance is $\dfrac{81}{13}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by $\sqrt{169}$ → $81$
• Divided by $169$ instead of by its square root → $\frac{81}{169}$

Answer: $(-2, -3)$

Complete the square in $x$ and in $y\mathclose{:}$ $\left(x + 2\right)^{2} + \left(y + 3\right)^{2} = 4 + 9 + 3 = 16\mathclose{.}$ So the centre is $(-2, -3)$ and the radius is $\sqrt{16} = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Copied the signs straight from the equation → $(2, 3)$
• Did not halve the coefficients → $(-4, -6)$

Answer: $7$

The centre is $(1, -2)$ and the radius $6\mathclose{.}$ The point is $\sqrt{5^2 + 12^2} = 13$ from the centre, so it lies outside the circle, and the nearest point of the circle is on the line through the centre: $|13 - 6| = 7\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the distance to the centre → $13$
• Added the radius instead of subtracting it → $19$

Answer: $y = - \frac{x}{4} - \frac{3}{2}$

The given line has slope $4\mathclose{,}$ so a perpendicular one has slope $-\dfrac{1}{4}\mathclose{.}$ Through $(-2, -1)\mathclose{:}$ $y = -\dfrac{1}{4}\left(x + 2\right) - 1\mathclose{,}$ which is $y = - \frac{x}{4} - \frac{3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the same slope, which gives a PARALLEL line → $y = 4 x + 7$
• Used $-4$ as the slope instead of $-\frac{1}{4}$ → $y = - 4 x - 9$

Answer: $60$

The slope is the tangent of that angle: $\tan\theta = \sqrt{3}\mathclose{,}$ so $\theta = 60^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the angle in radians → $\frac{\pi}{3}$
• Gave the angle with the $y$-axis → $30$

Answer: $6$

Complete the square in $x$ and in $y\mathclose{:}$ $\left(x + 4\right)^{2} + \left(y - 1\right)^{2} = 16 + 1 + 19 = 36\mathclose{.}$ So the centre is $(-4, 1)$ and the radius is $\sqrt{36} = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $r^2$ instead of $r$ → $36$
• Took the square root of the constant term alone → $\sqrt{19}$

Answer: $y = - \frac{15 x}{8} + \frac{103}{4}$

The radius to $P$ runs from the centre $(-5, -1)$ to $(10, 7)\mathclose{,}$ so its slope is $\dfrac{8}{15}\mathclose{.}$ The tangent is perpendicular to it, with slope $-\dfrac{15}{8}\mathclose{.}$ Through $P\mathclose{:}$ $y = -\dfrac{15}{8}\left(x - 10\right) + 7\mathclose{,}$ which is $y = - \frac{15 x}{8} + \frac{103}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the slope of the radius itself → $y = \frac{8 x}{15} + \frac{5}{3}$
• Turned the radius's slope over but kept its sign → $y = \frac{15 x}{8} - \frac{47}{4}$

Answer: $(5, 1)$

Substitute the line into the circle: a quadratic in $x\mathclose{,}$ whose roots are $x = 2$ and $x = 5\mathclose{.}$ The line then gives $y\mathclose{:}$ the points are $(2, -2)$ and $(5, 1)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other intersection point → $(2, -2)$
• Swapped the coordinates → $(1, 5)$

Answer: $17$

The horizontal gap is $|18 - 3| = 15$ and the vertical gap is $|-10 - (-2)| = 8\mathclose{,}$ so Pythagoras gives $\sqrt{15^2 + 8^2} = \sqrt{289} = 17\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two gaps instead of using Pythagoras → $23$
• Forgot the square root → $289$

Answer: $\dfrac{44}{5}$

For $ax + by + c = 0$ the distance is $\dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\mathclose{.}$ Here the top is $|3 \cdot (-6) + 4 \cdot (-4) - 10| = 44$ and the bottom $\sqrt{25} = 5\mathclose{,}$ so the distance is $\dfrac{44}{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by $\sqrt{25}$ → $44$
• Divided by $25$ instead of by its square root → $\frac{44}{25}$
• Left out the absolute value → $- \frac{44}{5}$

Answer: $\dfrac{1}{2}$

Perpendicular slopes multiply to $-1\mathclose{:}$ turn $-2$ over and change its sign, giving $\dfrac{1}{2}\mathclose{.}$ (Check: $-2 \times \dfrac{1}{2} = -1\mathclose{.})$

Common mistakes, and the answer each one gives:

• Turned the slope over but kept its sign → $- \frac{1}{2}$
• Changed the sign but did not turn the slope over → $2$

Answer: $30$

Each slope is the tangent of the line's angle with the $x$-axis: $0^\circ$ and $30^\circ\mathclose{.}$ The lines meet at the difference, $30^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the obtuse angle instead of the acute one → $150$

Answer: $\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 25$

The radius is the distance from the centre to the point: $\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5\mathclose{.}$ So the circle is $\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 25\mathclose{.}$

Common mistakes, and the answer each one gives:

• Put $r$ instead of $r^2$ on the right → $\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 5$
• Put the centre in with the wrong signs → $\left(x - 1\right)^{2} + \left(y + 1\right)^{2} = 25$

Answer: $(1, 4)$

Subtract one equation from the other: the squares cancel and a straight line is left — the common chord, $x = 1\mathclose{.}$ Put that into either circle: the points are $(1, 4)$ and $(1, 2)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other intersection point → $(1, 2)$
• Swapped the coordinates → $(4, 1)$