Question: Rewrite $x^2 + y^2 - 4x + 6y - 12 = 0$ in standard form and state the centre and radius.

Step 1. Group $x$-terms and $y$-terms; move the constant to RHS:

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

Step 2. Complete the square in each bracket.

• For $x$: add $\left(\tfrac{-4}{2}\right)^2 = 4$.
• For $y$: add $\left(\tfrac{6}{2}\right)^2 = 9$.

Balance the equation by adding both to the RHS too:

$$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9$$ $$(x - 2)^2 + (y + 3)^2 = 25$$

Step 3. Compare with $(x - a)^2 + (y - b)^2 = r^2$: Centre $= (2, -3)$, radius $= 5$.

⚠ Watch out: $(y + 3)^2 = (y - (-3))^2$, so the $y$-coordinate of the centre is $-3$, not $+3$. Track the sign.

Question: Find the equation of the circle with centre $(-1, 4)$ and radius $\sqrt{13}$, in general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

Step 1. Start in standard form:

$$(x - (-1))^2 + (y - 4)^2 = 13 \;\Rightarrow\; (x + 1)^2 + (y - 4)^2 = 13$$

Step 2. Expand:

$$\begin{aligned}x^2 + 2x + 1 + y^2 - 8y + 16 &= 13 \\ x^2 + y^2 + 2x - 8y + 4 &= 0\end{aligned}$$

Answer: $x^2 + y^2 + 2x - 8y + 4 = 0$ (so $g = 1$, $f = -4$, $c = 4$).

Tip: General-form coefficients give centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$. Quick cross-check: $g^2 + f^2 - c = 1 + 16 - 4 = 13$ ✓.

Question: A circle lies entirely in the first quadrant and touches both the $x$-axis and the $y$-axis. Given that it passes through the point $(1, 2)$, find the equation of each possible circle.

Step 1. Touching an axis means the distance from the centre to that axis equals the radius. Touching both axes (first quadrant) forces the centre to be $(r, r)$ with $r > 0$:

$$(x - r)^2 + (y - r)^2 = r^2$$

Step 2. The circle passes through $(1, 2)$ — substitute:

$$(1 - r)^2 + (2 - r)^2 = r^2$$ $$1 - 2r + r^2 + 4 - 4r + r^2 = r^2$$ $$r^2 - 6r + 5 = 0$$

Step 3. Factorise and solve:

$$(r - 1)(r - 5) = 0 \implies r = 1 \;\text{ or }\; r = 5$$

Both roots are positive, so both give a valid circle.

Step 4. State both equations:

• $r = 1$: centre $(1, 1)$, $\;(x - 1)^2 + (y - 1)^2 = 1$
• $r = 5$: centre $(5, 5)$, $\;(x - 5)^2 + (y - 5)^2 = 25$

Check the second: $(1 - 5)^2 + (2 - 5)^2 = 16 + 9 = 25 = r^2$ ✓

⚠ Watch out: Do not discard $r = 5$ because it "looks too big" — reject a root only for a real reason (e.g. $r \le 0$). Here a small circle and a large circle both touch the axes and pass through $(1, 2)$.

Question: Find the equation of the circle passing through $A(0, 0)$, $B(8, 0)$, $C(4, 4)$.

Step 1. Assume general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

Step 2. Substitute each point:

• $A(0, 0)$: $\;0 + 0 + 0 + 0 + c = 0 \Rightarrow c = 0$
• $B(8, 0)$: $\;64 + 16g + c = 0 \Rightarrow 16g = -64 \Rightarrow g = -4$
• $C(4, 4)$: $\;16 + 16 + 8g + 8f + c = 0 \Rightarrow 32 - 32 + 8f = 0 \Rightarrow f = 0$

Step 3. Equation: $x^2 + y^2 - 8x = 0$.

Step 4. Convert: $(x - 4)^2 + y^2 = 16$. Centre $(4, 0)$, radius $4$.

Tip: Always sub $A$ first if it's $(0, 0)$ — gives $c$ immediately.

Question: Find the equation of the circle with diameter endpoints $P(2, 1)$ and $Q(8, 9)$.

Step 1. The centre is the midpoint of $PQ$:

$$C = \left(\tfrac{2 + 8}{2}, \tfrac{1 + 9}{2}\right) = (5, 5)$$

Step 2. The radius is half the diameter $|PQ|$:

$$|PQ| = \sqrt{(8-2)^2 + (9-1)^2} = \sqrt{36 + 64} = 10$$ $$r = \tfrac{10}{2} = 5$$

Step 3. Standard form:

$$(x - 5)^2 + (y - 5)^2 = 25$$

Tip: "Diameter endpoints" is the simplest three-points case — only two points given, and the geometry handles the rest.

Question: Find $k$ such that the line $y = 2x + k$ is tangent to the circle $x^2 + y^2 - 4x + 6y + 4 = 0$.

Step 1. Substitute the line into the circle:

$$x^2 + (2x + k)^2 - 4x + 6(2x + k) + 4 = 0$$

Step 2. Expand and collect:

$$\begin{aligned}x^2 + 4x^2 + 4kx + k^2 - 4x + 12x + 6k + 4 &= 0 \\ 5x^2 + (4k + 8)x + (k^2 + 6k + 4) &= 0\end{aligned}$$

Step 3. Tangency $\Leftrightarrow$ discriminant $= 0$:

$$(4k + 8)^2 - 4(5)(k^2 + 6k + 4) = 0$$ $$16k^2 + 64k + 64 - 20k^2 - 120k - 80 = 0$$ $$-4k^2 - 56k - 16 = 0 \;\Rightarrow\; k^2 + 14k + 4 = 0$$

Step 4. Solve via quadratic formula:

$$k = \dfrac{-14 \pm \sqrt{196 - 16}}{2} = \dfrac{-14 \pm \sqrt{180}}{2} = -7 \pm 3\sqrt{5}$$

Answer: $k = -7 + 3\sqrt{5}$ or $k = -7 - 3\sqrt{5}$ — there are two tangent lines (one above and one below the circle).

Question: Is the line $3x + 4y = 25$ tangent to the circle $x^2 + y^2 = 25$?

Method 1: Discriminant.

Step 1. From the line: $y = \tfrac{25 - 3x}{4}$. Substitute into the circle:

$$x^2 + \left(\tfrac{25 - 3x}{4}\right)^2 = 25$$

Multiply by 16:

$$16x^2 + (25 - 3x)^2 = 400$$ $$16x^2 + 625 - 150x + 9x^2 = 400$$ $$25x^2 - 150x + 225 = 0 \;\Rightarrow\; x^2 - 6x + 9 = 0$$

Step 2. Discriminant: $\Delta = 36 - 36 = 0$. ✓ Tangent.

Method 2: Distance from centre = radius.

Centre $(0, 0)$. Distance from $(0, 0)$ to line $3x + 4y - 25 = 0$:

$$d = \dfrac{|3(0) + 4(0) - 25|}{\sqrt{9 + 16}} = \dfrac{25}{5} = 5$$

Radius $= \sqrt{25} = 5$. Distance $=$ radius $\Rightarrow$ tangent. ✓

Tip: Method 2 (distance from centre) is often quicker if both the centre and the line equation $ax + by + c = 0$ are easy to read.

Question: Find the equation of the tangent to $(x - 1)^2 + (y - 2)^2 = 25$ at the point $T(4, 6)$.

Step 1. The tangent is perpendicular to the radius at the point of tangency.

Gradient of radius $CT$:

$$m_{CT} = \dfrac{6 - 2}{4 - 1} = \dfrac{4}{3}$$

Step 2. Tangent gradient (negative reciprocal):

$$m_{\text{tan}} = -\dfrac{3}{4}$$

Step 3. Equation of tangent through $T(4, 6)$:

$$y - 6 = -\tfrac{3}{4}(x - 4) \;\Rightarrow\; y = -\tfrac{3}{4}x + 9$$

Answer: $y = -\tfrac{3}{4}x + 9$ (or $3x + 4y = 36$).

Why this works: Radius ⊥ tangent at the point of tangency. So once you know the gradient of the radius, the perpendicular gives the tangent gradient in one step.

Question: A circle has diameter endpoints $A(0, 0)$ and $B(6, 8)$. Find the equation of the tangent at $A$.

Step 1. The tangent at $A$ is perpendicular to the radius (or equivalently the diameter) at $A$.

Gradient of $AB$:

$$m_{AB} = \dfrac{8 - 0}{6 - 0} = \dfrac{4}{3}$$

Step 2. Perpendicular gradient: $m_{\text{tan}} = -\tfrac{3}{4}$.

Step 3. Tangent through $A(0, 0)$:

$$y - 0 = -\tfrac{3}{4}(x - 0) \;\Rightarrow\; y = -\tfrac{3}{4}x$$

Answer: $y = -\tfrac{3}{4}x$ (or $3x + 4y = 0$).

Tip: "Tangent at an endpoint of a diameter" never needs the centre — just the gradient of the diameter, then $\perp$.

Question: A line $y = x + 1$ cuts the circle $x^2 + y^2 = 25$. Find the length of the chord.

Step 1. Distance from centre $(0, 0)$ to line $x - y + 1 = 0$:

$$d = \dfrac{|0 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \dfrac{1}{\sqrt{2}}$$

Step 2. Half-chord length by Pythagoras (in the right triangle formed by radius, perpendicular distance, and half-chord):

$$\text{(half-chord)}^2 = r^2 - d^2 = 25 - \tfrac{1}{2} = \tfrac{49}{2}$$ $$\text{half-chord} = \sqrt{\tfrac{49}{2}} = \tfrac{7}{\sqrt{2}}$$

Step 3. Full chord length:

$$2 \times \tfrac{7}{\sqrt{2}} = \tfrac{14}{\sqrt{2}} = 7\sqrt{2}$$

Answer: Chord length $= 7\sqrt{2}$ units.

Why this is faster: Solving for intersection points then computing distance works, but the perpendicular-distance method avoids the messy quadratic — direct Pythagoras.

Question: Find the points where $y = x - 1$ meets the circle $x^2 + y^2 = 13$.

Step 1. Substitute the line into the circle:

$$x^2 + (x - 1)^2 = 13$$

Step 2. Expand and solve:

$$\begin{aligned}x^2 + x^2 - 2x + 1 &= 13 \\ 2x^2 - 2x - 12 &= 0 \\ x^2 - x - 6 &= 0 \\ (x - 3)(x + 2) &= 0\end{aligned}$$

So $x = 3$ or $x = -2$.

Step 3. Get the $y$-coordinates from $y = x - 1$:

• $x = 3 \Rightarrow y = 2$
• $x = -2 \Rightarrow y = -3$

Answer: Intersection points are $(3, 2)$ and $(-2, -3)$.

Tip: Always sub the line into the circle (not vice versa). The line is linear, so the substitution yields one quadratic in $x$ — clean.

Question: Show that the circles $x^2 + y^2 = 4$ and $(x - 5)^2 + y^2 = 9$ are externally tangent.

Step 1. Identify centres and radii:

• $C_1 = (0, 0)$, $r_1 = 2$
• $C_2 = (5, 0)$, $r_2 = 3$

Step 2. Distance between centres:

$$d = \sqrt{(5 - 0)^2 + (0 - 0)^2} = 5$$

Step 3. Compare with sum / difference of radii:

• Sum: $r_1 + r_2 = 2 + 3 = 5$ ✓
• Difference: $|r_1 - r_2| = 1$

Since $d = r_1 + r_2$, the circles are externally tangent.

Three configurations to remember:

• $d > r_1 + r_2$: circles don't meet (separate)
• $d = r_1 + r_2$: externally tangent
• $|r_1 - r_2| < d < r_1 + r_2$: circles intersect at 2 points
• $d = |r_1 - r_2|$: internally tangent
• $d < |r_1 - r_2|$: one circle inside the other (no contact)

Question: Find the equation of the common chord of the circles $x^2 + y^2 - 4x - 2y - 4 = 0 \quad (C_1)$ $x^2 + y^2 - 8x - 6y + 16 = 0 \quad (C_2)$

Step 1. Subtract $C_2$ from $C_1$ (the $x^2 + y^2$ terms cancel):

$$(C_1) - (C_2): \;(-4x - 2y - 4) - (-8x - 6y + 16) = 0$$ $$4x + 4y - 20 = 0 \;\Rightarrow\; x + y = 5$$

Answer: Common chord: $x + y = 5$.

Why this works: A point on both circles satisfies both equations. Subtracting eliminates the quadratic terms, leaving a linear equation — that's the line through both intersection points (the common chord).

Tip: Use this trick to find the common chord WITHOUT having to compute the intersection points themselves.

Question: $A(0, 0)$ and $B(10, 0)$ are endpoints of a diameter of a circle. Show that $C(8, 4)$ lies on the circle, and verify $\angle ACB = 90°$.

Step 1. Centre of circle = midpoint of $AB$:

$$M = \left(\tfrac{0 + 10}{2}, \tfrac{0 + 0}{2}\right) = (5, 0)$$

Radius $= \tfrac{1}{2}|AB| = 5$.

Step 2. Check $C(8, 4)$ on circle:

|MC| = \sqrt{(8 - 5)^2 + (4 - 0)^2} = \sqrt{9 + 16} = 5 $$ ✓ **Step 3.** Verify $\angle ACB = 90°$ via $\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$: $\overrightarrow{CA} = (0 - 8, 0 - 4) = (-8, -4)$ $\overrightarrow{CB} = (10 - 8, 0 - 4) = (2, -4)$ $\overrightarrow{CA} \cdot \overrightarrow{CB} = (-8)(2) + (-4)(-4) = -16 + 16 = 0$ ✓ So $\angle ACB = 90°$ — confirming the angle-in-semicircle theorem. **Tip:** Two equivalent tests for "right angle at $C$": - Distance test: $|MC| = r$ (C lies on the circle, so the theorem applies) - Vector test: $\overrightarrow{CA} \cdot \overrightarrow{CB} = 0$

Question: A chord of the circle $x^2 + y^2 = 25$ has midpoint $M(3, 1)$. Find the equation of the chord.

Step 1. Theorem: the line from the centre to the midpoint of a chord is perpendicular to the chord.

Step 2. Gradient of line from centre $(0, 0)$ to $M(3, 1)$:

$$m_{OM} = \tfrac{1 - 0}{3 - 0} = \tfrac{1}{3}$$

Step 3. Chord's gradient (perpendicular):

$$m_{\text{chord}} = -3$$

Step 4. Equation of chord through $M(3, 1)$:

$$y - 1 = -3(x - 3) \;\Rightarrow\; y = -3x + 10$$

Answer: Chord: $y = -3x + 10$ (or $3x + y = 10$).

Why this works: The radius drawn to the midpoint of any chord is perpendicular to that chord — a classical Euclidean result. Quick consequence: if you know the midpoint, you know everything about the chord's direction.

Question: Determine whether each point lies inside, on, or outside the circle $(x - 2)^2 + (y + 1)^2 = 25$: (a) $P(5, 3)$ (b) $Q(7, 2)$ (c) $R(0, 6)$

Method: Compute $(x - 2)^2 + (y + 1)^2$ and compare with $r^2 = 25$.

• $< 25$: inside
• $= 25$: on
• $> 25$: outside

Step 1. $P(5, 3)$: $(5 - 2)^2 + (3 + 1)^2 = 9 + 16 = 25$ ✓ on the circle.

Step 2. $Q(7, 2)$: $(7 - 2)^2 + (2 + 1)^2 = 25 + 9 = 34 > 25$ ⟹ outside.

Step 3. $R(0, 6)$: $(0 - 2)^2 + (6 + 1)^2 = 4 + 49 = 53 > 25$ ⟹ outside.

Tip: This is just the distance test in disguise. $\sqrt{(x-a)^2 + (y-b)^2}$ is the distance from $(x, y)$ to the centre — compare with the radius. Squaring both sides skips the square root.

Question: Find the values of $k$ for which the point $(k, 2)$ lies inside the circle $x^2 + y^2 - 4x - 2y - 11 = 0$.

Step 1. Convert to standard form by completing the square:

$$(x - 2)^2 + (y - 1)^2 = 16$$

Centre $(2, 1)$, radius $4$.

Step 2. "Inside" means $(k - 2)^2 + (2 - 1)^2 < 16$:

$$(k - 2)^2 + 1 < 16$$ $$(k - 2)^2 < 15$$ $$|k - 2| < \sqrt{15}$$ $$2 - \sqrt{15} < k < 2 + \sqrt{15}$$

Answer: $2 - \sqrt{15} < k < 2 + \sqrt{15}$.

⚠ Watch out: Strict inequalities ($<$, not $\leq$) — "inside" doesn't include the boundary. Use $\leq$ if the question says "inside or on."