Coordinate Geometry — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for coordinate geometry.

UT-1: Circle Equation by Completing the Square

Question:

Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.

Solution:

Complete the square for $x$ and $y$:

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

$(x - 3)^2 + (y + 2)^2 = 25$

Centre: $(3,\; -2)$Radius: $5$.

UT-2: Perpendicular Slopes Product Equals -1

Question:

The line $L_1$ passes through $A(1, 3)$ and $B(5, 7)$. The line $L_2$ passes through $C(4, -1)$ and is perpendicular to $L_1$. Find the equation of $L_2$.

Solution:

Slope of $L_1$: $m_1 = \dfrac{7 - 3}{5 - 1} = \dfrac{4}{4} = 1$.

Slope of $L_2$: $m_2 = \dfrac{-1}{m_1} = -1$.

Using point-slope form with point $C(4, -1)$:

$y - (-1) = -1(x - 4)$

$y + 1 = -x + 4$

$y = -x + 3$

Or $x + y - 3 = 0$.

UT-3: Circle Tangent from External Point

Question:

Find the length of the tangent from the point $P(8, 3)$ to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$.

Solution:

Complete the square: $(x - 2)^2 + (y - 3)^2 = 4$.

Centre $C = (2, 3)$Radius $r = 2$.

Distance from $P$ to $C$: $PC = \sqrt{(8-2)^2 + (3-3)^2} = \sqrt{36} = 6$.

Length of tangent: $PT = \sqrt{PC^2 - r^2} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$.

UT-4: Locus Definition

Question:

A point $P(x, y)$ moves such that its distance from $(3, -1)$ is always twice its distance from the line $x = -1$. Find the equation of the locus of $P$.

Solution:

Distance from $P$ to $(3, -1)$: $\sqrt{(x-3)^2 + (y+1)^2}$.

Distance from $P$ to $x = -1$: $|x + 1|$.

$\sqrt{(x-3)^2 + (y+1)^2} = 2|x + 1|$

Square both sides:

$(x-3)^2 + (y+1)^2 = 4(x+1)^2$

$x^2 - 6x + 9 + y^2 + 2y + 1 = 4x^2 + 8x + 4$

$-3x^2 - 14x + y^2 + 2y + 6 = 0$

$3x^2 - y^2 + 14x - 2y - 6 = 0$

This is a hyperbola.

UT-5: Intersection of Line and Circle

Question:

Find the coordinates of the points of intersection of the line $y = 2x - 1$ and the circle $x^2 + y^2 - 10x - 6y + 25 = 0$.

Solution:

Substitute $y = 2x - 1$ into the circle equation:

$x^2 + (2x - 1)^2 - 10x - 6(2x - 1) + 25 = 0$

$x^2 + 4x^2 - 4x + 1 - 10x - 12x + 6 + 25 = 0$

$5x^2 - 26x + 32 = 0$

$(5x - 16)(x - 2) = 0$

$x = \frac{16}{5} \quad \text{or} \quad x = 2$

$x = \dfrac{16}{5}$: $y = \dfrac{32}{5} - 1 = \dfrac{27}{5}$. Point: $\left(\dfrac{16}{5},\; \dfrac{27}{5}\right)$.

$x = 2$: $y = 3$. Point: $(2,\; 3)$.

Integration Tests

Tests synthesis of coordinate geometry with other topics.

IT-1: Coordinate Geometry and Quadratics (with Quadratics)

Question:

The line $y = mx + c$ is tangent to the circle $x^2 + y^2 = 4$. Express $c$ in terms of $m$.

Solution:

Substitute $y = mx + c$:

$x^2 + (mx + c)^2 = 4$

$x^2 + m^2x^2 + 2mcx + c^2 - 4 = 0$

$(1 + m^2)x^2 + 2mcx + (c^2 - 4) = 0$

For tangency, $\Delta = 0$:

$(2mc)^2 - 4(1 + m^2)(c^2 - 4) = 0$

$4m^2c^2 - 4(1 + m^2)(c^2 - 4) = 0$

$m^2c^2 - (1 + m^2)c^2 + 4(1 + m^2) = 0$

$m^2c^2 - c^2 - m^2c^2 + 4 + 4m^2 = 0$

$-c^2 + 4 + 4m^2 = 0$

$c^2 = 4(1 + m^2)$

$c = \pm 2\sqrt{1 + m^2}$

IT-2: Coordinate Geometry and Trigonometry (with Trigonometry)

Question:

The points $A(1, 0)$ and $B(0, 1)$ are on the unit circle. Find the coordinates of point $C$ on the unit circle such that triangle $ABC$ is equilateral.

Solution:

$OA = OB = OC = 1$ (all on unit circle).

Angle $AOB$: $\angle AOB$ where $A = (1, 0)$ and $B = (0, 1)$.

$\cos \angle AOB = \dfrac{\vec{OA} \cdot \vec{OB}}{|OA||OB|} = \dfrac{0 + 0}{1} = 0$So $\angle AOB = 90°$.

For an equilateral triangle inscribed in a circle, each angle is $60°$. But $\angle AOB = 90° \neq 60°$So there is no equilateral triangle with vertices $A$, $B$, $C$ all on the unit circle.

This tests the misconception of assuming any three points on a circle can form an equilateral triangle.

IT-3: Coordinate Geometry and Functions (with Functions)

Question:

The parabola $y = x^2 - 4x + 7$ has vertex $V$. Find the coordinates of $V$And the equation of the axis of symmetry. If the parabola is reflected in the $y$-axis, find the equation of the reflected parabola.

Solution:

$y = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$.

Vertex: $V = (2, 3)$.

Axis of symmetry: $x = 2$.

Reflecting in the $y$-axis replaces $x$ with $-x$:

$y = (-x)^2 - 4(-x) + 7 = x^2 + 4x + 7 = (x + 2)^2 + 3$.

The reflected parabola has vertex $(-2, 3)$ and axis $x = -2$.

Worked Examples

WE-1: Distance and Midpoint Verification

Question:

$A(3, -1)$, $B(7, 5)$And $C(11, -1)$ are three points.

(a) Show that triangle $ABC$ is isosceles. (2 marks) (b) Find the area of triangle $ABC$. (3 marks)

Solution:

(a) $AB = \sqrt{(7-3)^2 + (5-(-1))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

$BC = \sqrt{(11-7)^2 + (-1-5)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

$AC = \sqrt{(11-3)^2 + (-1-(-1))^2} = \sqrt{64} = 8$.

Since $AB = BC$Triangle $ABC$ is isosceles.

(b) Base $AC = 8$. Height is the perpendicular from $B$ to $AC$.

Since $AC$ is horizontal ($y = -1$), the height $= |5 - (-1)| = 6$.

Area $= \dfrac{1}{2} \times 8 \times 6 = 24$ square units.

WE-2: Equation of Circle Through Two Points

Question:

Find the equation of the circle with centre on the line $y = 2x$ that passes through the points $(1, 3)$ and $(4, 6)$.

Solution:

Let the centre be $C(a, 2a)$.

$CA = CB$ (equal radii):

$\sqrt{(a-1)^2 + (2a-3)^2} = \sqrt{(a-4)^2 + (2a-6)^2}$

Square both sides:

$(a-1)^2 + (2a-3)^2 = (a-4)^2 + (2a-6)^2$

$a^2 - 2a + 1 + 4a^2 - 12a + 9 = a^2 - 8a + 16 + 4a^2 - 24a + 36$

$5a^2 - 14a + 10 = 5a^2 - 32a + 52$

$18a = 42$

$a = \frac{42}{18} = \frac{7}{3}$

Centre: $\left(\dfrac{7}{3},\; \dfrac{14}{3}\right)$.

Radius squared: $\left(\dfrac{7}{3} - 1\right)^2 + \left(\dfrac{14}{3} - 3\right)^2 = \dfrac{16}{9} + \dfrac{25}{9} = \dfrac{41}{9}$.

Equation: $\left(x - \dfrac{7}{3}\right)^2 + \left(y - \dfrac{14}{3}\right)^2 = \dfrac{41}{9}$Or $9x^2 + 9y^2 - 42x - 84y + 128 = 0$.

WE-3: Locus of Moving Point

Question:

A point $P(x, y)$ moves so that its distance from the point $(2, -3)$ always equals its distance from the line $y = 1$. Find the equation of the locus of $P$.

Solution:

Distance from $P$ to $(2, -3)$: $\sqrt{(x-2)^2 + (y+3)^2}$.

Distance from $P$ to $y = 1$: $|y - 1|$.

$\sqrt{(x-2)^2 + (y+3)^2} = |y - 1|$

Squaring: $(x-2)^2 + (y+3)^2 = (y-1)^2$.

$x^2 - 4x + 4 + y^2 + 6y + 9 = y^2 - 2y + 1$

$x^2 - 4x + 8y + 12 = 0$

$x^2 - 4x + 8y + 12 = 0$

This is a parabola (since it has one squared term and one linear term in the other variable).

WE-4: Angle Between Two Lines

Question:

Find the acute angle between the lines $3x - y + 2 = 0$ and $x + 2y - 7 = 0$.

Solution:

Line 1: $3x - y + 2 = 0 \implies y = 3x + 2$Slope $m_1 = 3$.

Line 2: $x + 2y - 7 = 0 \implies y = -\dfrac{1}{2}x + \dfrac{7}{2}$Slope $m_2 = -\dfrac{1}{2}$.

$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{3 - (-1/2)}{1 + 3(-1/2)}\right| = \left|\frac{7/2}{1 - 3/2}\right| = \left|\frac{7/2}{-1/2}\right| = 7$

$\theta = \arctan(7) \approx 81.9°$

WE-5: Finding the Point of Intersection

Question:

Find the coordinates of the point of intersection of the lines $2x + 3y = 13$ and $4x - y = 5$.

Solution:

From the second equation: $y = 4x - 5$.

Substitute into the first: $2x + 3(4x - 5) = 13$.

$2x + 12x - 15 = 13$

$14x = 28 \implies x = 2$

$y = 4(2) - 5 = 3$

Point of intersection: $(2, 3)$.

WE-6: Shortest Distance from Point to Line

Question:

Find the shortest distance from the point $(3, -1)$ to the line $4x + 3y - 10 = 0$.

Solution:

Using the point-to-line distance formula:

$d = \frac{|4(3) + 3(-1) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|12 - 3 - 10|}{5} = \frac{1}{5}$

WE-7: Two Circles Touching Externally

Question:

Circle $C_1$: $x^2 + y^2 - 4x - 6y + 9 = 0$. Circle $C_2$: $x^2 + y^2 + 6x + 2y - 15 = 0$.

(a) Find the centres and radii of $C_1$ and $C_2$. (3 marks) (b) Show that $C_1$ and $C_2$ touch externally. (2 marks)

Solution:

(a) $C_1$: $(x-2)^2 + (y-3)^2 = 4$. Centre $O_1 = (2, 3)$, $r_1 = 2$.

$C_2$: $(x+3)^2 + (y+1)^2 = 25$. Centre $O_2 = (-3, -1)$, $r_2 = 5$.

(b) Distance between centres:

$O_1 O_2 = \sqrt{(2-(-3))^2 + (3-(-1))^2} = \sqrt{25 + 16} = \sqrt{41}$

Sum of radii: $r_1 + r_2 = 2 + 5 = 7$.

$\sqrt{41} \neq 7$ ($\sqrt{41} \approx 6.4$), so $C_1$ and $C_2$ do NOT touch externally.

In fact, $O_1 O_2 = \sqrt{41} < 7 = r_1 + r_2$So the circles overlap (intersect at two points).

WE-8: Equation of Perpendicular Bisector

Question:

Find the equation of the perpendicular bisector of the segment joining $A(-2, 5)$ and $B(4, -1)$.

Solution:

Midpoint: $M = \left(\dfrac{-2+4}{2},\; \dfrac{5+(-1)}{2}\right) = (1, 2)$.

Slope of $AB$: $m = \dfrac{-1-5}{4-(-2)} = \dfrac{-6}{6} = -1$.

Slope of perpendicular bisector: $m_{\perp} = 1$.

Equation: $y - 2 = 1(x - 1) \implies y = x + 1$Or $x - y + 1 = 0$.

Common Pitfalls

1. Incorrect sign when completing the square in circle equations. When completing the square for $x^2 - 6x$The result is $(x - 3)^2 - 9$Not $(x - 3)^2 + 9$. Remember: you add and subtract the same quantity. Always verify by expanding back.

2. Forgetting to square the distance formula. When equating distances (e.g. For a locus problem), square both sides immediately to avoid messy square roots. A common error is trying to work with $\sqrt{\ldots} = \sqrt{\ldots}$ without squaring.

3. Confusing the radius with the diameter in circle equations. The standard form $(x - h)^2 + (y - k)^2 = r^2$ gives the radius squared on the right side. If the equation is $(x-h)^2 + (y-k)^2 = d^2$Then $d$ is the diameter, not the radius.

4. Wrong slope for perpendicular lines. If a line has slope $m$The perpendicular line has slope $-\dfrac{1}{m}$Not $\dfrac{1}{m}$. This sign error is extremely common in DSE.

5. Not considering both intersection points of line and circle. When a line intersects a circle, there can be 0, 1 (tangent), or 2 intersection points. Always check the discriminant of the resulting quadratic.

DSE Exam-Style Questions

DSE-1

The equation of a circle $C$ is $x^2 + y^2 + 4x - 10y + 20 = 0$.

(a) Find the centre and radius of $C$. (3 marks) (b) Find the equation of the tangent to $C$ at the point $(-1, 5)$. (4 marks) (c) The tangent in (b) meets the $y$-axis at $P$. Find the coordinates of $P$. (2 marks)

Solution:

(a) $(x^2 + 4x + 4) + (y^2 - 10y + 25) = -20 + 4 + 25 = 9$.

$(x + 2)^2 + (y - 5)^2 = 9$.

Centre $= (-2, 5)$Radius $= 3$.

(b) The tangent at $(-1, 5)$ is perpendicular to the radius from $(-2, 5)$ to $(-1, 5)$.

Slope of radius $= \dfrac{5 - 5}{-1 - (-2)} = 0$ (horizontal).

So the tangent is vertical: $x = -1$.

(c) $x = -1$ meets the $y$-axis ($x = 0$)? A vertical line $x = -1$ is parallel to the $y$-axis and never meets it. There is no intersection point $P$.

This reveals an important check: when the point $(-1, 5)$ has the same $y$-coordinate as the centre, the radius is horizontal and the tangent is vertical.

DSE-2

The line $L: y = mx + 1$ intersects the circle $x^2 + y^2 = 9$ at two distinct points.

(a) Show that $(1 + m^2)x^2 + 2mx - 8 = 0$. (2 marks) (b) Find the range of values of $m$ for which $L$ intersects the circle at two distinct points. (3 marks) (c) For $m = 1$Find the length of the chord of intersection. (3 marks)

Solution:

(a) Substitute $y = mx + 1$ into $x^2 + y^2 = 9$:

$x^2 + (mx + 1)^2 = 9$

$x^2 + m^2 x^2 + 2mx + 1 = 9$

$(1 + m^2)x^2 + 2mx - 8 = 0$.

(b) Two distinct intersections require $\Delta > 0$:

$\Delta = (2m)^2 - 4(1+m^2)(-8) = 4m^2 + 32(1 + m^2) = 4m^2 + 32 + 32m^2 = 36m^2 + 32$.

Since $36m^2 + 32 > 0$ for all real $m$The line always intersects the circle at two distinct points for all $m$.

(c) For $m = 1$: $(1 + 1)x^2 + 2x - 8 = 0 \implies 2x^2 + 2x - 8 = 0 \implies x^2 + x - 4 = 0$.

$x = \dfrac{-1 \pm \sqrt{1 + 16}}{2} = \dfrac{-1 \pm \sqrt{17}}{2}$.

$y$-coordinates: $y = x + 1$So $y = \dfrac{1 \pm \sqrt{17}}{2}$.

Chord length $= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(\sqrt{17})^2 + (\sqrt{17})^2} = \sqrt{34}$.

DSE-3

A variable point $P(x, y)$ moves such that its distance from $A(1, 2)$ is always equal to its distance from $B(5, 6)$.

(a) Find the equation of the locus of $P$. (3 marks) (b) Verify that $(3, 4)$ lies on the locus. (1 mark) (c) Describe the locus geometrically. (1 mark)

Solution:

(a) $\sqrt{(x-1)^2 + (y-2)^2} = \sqrt{(x-5)^2 + (y-6)^2}$.

Squaring: $(x-1)^2 + (y-2)^2 = (x-5)^2 + (y-6)^2$.

$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 10x + 25 + y^2 - 12y + 36$.

$-2x - 4y + 5 = -10x - 12y + 61$.

$8x + 8y - 56 = 0$.

$x + y - 7 = 0$.

(b) Check $(3, 4)$: $3 + 4 - 7 = 0$. Yes, it lies on the locus.

(c) The locus is the perpendicular bisector of the segment $AB$Which is a straight line.

DSE-4

The vertices of a triangle are $A(0, 0)$, $B(8, 0)$And $C(4, 6)$.

(a) Find the equation of the median from $C$ to $AB$. (2 marks) (b) Find the equation of the altitude from $A$ to $BC$. (3 marks) (c) Find the coordinates of the intersection of the median and the altitude. (3 marks)

Solution:

(a) Midpoint of $AB$: $M = (4, 0)$.

Median from $C(4, 6)$ to $M(4, 0)$: this is a vertical line $x = 4$.

(b) Slope of $BC$: $m_{BC} = \dfrac{0 - 6}{8 - 4} = \dfrac{-6}{4} = -\dfrac{3}{2}$.

Slope of altitude from $A$: $m_{\perp} = \dfrac{2}{3}$.

Equation: $y - 0 = \dfrac{2}{3}(x - 0) \implies y = \dfrac{2}{3}x$Or $2x - 3y = 0$.

(c) Intersection of $x = 4$ and $2x - 3y = 0$: $2(4) - 3y = 0 \implies y = \dfrac{8}{3}$.

Intersection point: $\left(4,\; \dfrac{8}{3}\right)$.

DSE-5

$C_1: (x-1)^2 + (y+2)^2 = 25$ and $C_2: (x-7)^2 + (y-4)^2 = 9$.

(a) Find the distance between the centres of $C_1$ and $C_2$. (2 marks) (b) Determine the number of intersection points of $C_1$ and $C_2$. (2 marks) (c) Find the equation of the common chord of $C_1$ and $C_2$. (4 marks)

Solution:

(a) Centres: $O_1 = (1, -2)$, $O_2 = (7, 4)$.

$O_1 O_2 = \sqrt{(7-1)^2 + (4-(-2))^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$.

(b) $r_1 = 5$, $r_2 = 3$.

$|r_1 - r_2| = 2$, $r_1 + r_2 = 8$.

$2 < 6\sqrt{2} < 8$ (since $6\sqrt{2} \approx 8.49$), so actually $O_1 O_2 > r_1 + r_2$.

The circles do not intersect. There are 0 intersection points.

Wait: $6\sqrt{2} = 6 \times 1.414 = 8.485 > 8 = r_1 + r_2$.

Therefore the circles are separate (0 intersection points). Part (c) would have no common chord. The question may have an error, or the circles may need adjustment. If the problem is solvable, we proceed assuming the circles do intersect.