Origin-centred circles and tangents

(a) $7$

1. $x^2 + y^2 = r^2$, so $r^2 = 49$ and $r = 7$.

• B1 The correct answer, $7$.

(b) Outside: $5^2 + 5^2 = 50 > 49$.

1. Distance from the centre squared: $5^2 + 5^2 = 50$.
2. $50 > 49$, so the point is outside the circle.

• M1 Working out $5^2 + 5^2 = 50$.
• C1 Outside, because $50 > 49$.

(a) $x^2 + y^2 = 100$

1. $r^2 = (-6)^2 + 8^2 = 36 + 64 = 100$.

• M1 Finding $r^2 = 100$ (or $r = 10$).
• A1 $x^2 + y^2 = 100$.

(b) $y = 8$ or $y = -8$

1. $36 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = \pm 8$.

• B1 The correct answer, $8$ and $-8$.

(a) $7$

1. $$58-3^{2}$$
2. Substitute x = 3: 9 + $y^{2}$ = 58, so $y^{2}$ = 49.
3. The square roots are ±7; use the positive value y = 7.

• P1 Establishing $58-3^{2}$ or an equivalent valid method.
• A1 Correct answer: $7$

(a) $5x + 12y = 169$

1. $$12/5$$
2. $$y-12=(-5/12)(x-5)$$
3. The radius from (0, 0) to (5, 12) has gradient 12/5.
4. A tangent is perpendicular to the radius, so its gradient is $-5$/12.
5. Using the point (5, 12), y $-$ 12 = $-$(5/12)(x $-$ 5).
6. Multiply by 12 and rearrange: 5x + 12y = 169.

• P1 Establishing $12/5$ or an equivalent valid method.
• P1 Establishing $y-12=(-5/12)(x-5)$ or an equivalent valid method.
• A1 Correct answer: $5x + 12y = 169$

(a) $(-9,12)$ and $(9,12)$

1. Substitute the fixed y-coordinate into the circle.
   $$x^{2}=225-12^{2}$$
2. Take both signs of the square root.
   $$x^{2}=81$$
3. Therefore $(-9,12)$ and $(9,12)$.

• M1 Substitute the fixed y-coordinate into the circle.
• M1 Take both signs of the square root.
• A1 Correct answer: $(-9,12)$ and $(9,12)$

(b) $18$

1. Subtract their x-coordinates because the y-coordinates are equal.
   $$9-(-9)$$
2. Therefore $18$.

• M1 Subtract their x-coordinates because the y-coordinates are equal.
• A1 Correct answer: $18$

(a) $(12,16)$

1. The perpendicular radius has gradient 4/3, so write y=4x/3.
   $$y=4x/3$$
2. Substitute in the circle and choose the positive x-coordinate.
   $$x^{2}+(4x/3)^{2}=400$$
3. Therefore $(12,16)$.

• M1 The perpendicular radius has gradient 4/3, so write y=4x/3.
• M1 Substitute in the circle and choose the positive x-coordinate.
• A1 Correct answer: $(12,16)$

(b) $25$

1. Substitute the contact point into y=−3x/4+c.
   $$16+3\times 12/4$$
2. Therefore $25$.

• M1 Substitute the contact point into y=−3x/4+c.
• A1 Correct answer: $25$

(a) $25$ square units

1. Radius gradient and tangent gradient.
   $$\frac{-2}{4} = -\frac{1}{2} \Rightarrow m = 2$$
2. Tangent through $(4, -2)$.
   $$-2 = 2 \times 4 + c \Rightarrow c = -10 \Rightarrow y = 2x - 10$$
3. Intercepts.
   $$A = (5, 0), \qquad B = (0, -10)$$
4. Area.
   $$\frac{1}{2} \times 5 \times 10 = 25$$

• P1 Finding the tangent gradient 2 from the radius gradient $-\frac{1}{2}$.
• P1 Finding the tangent equation $y = 2x - 10$.
• P1 Finding both intercepts: $(5, 0)$ and $(0, -10)$.
• A1 The correct answer, $25$.

(a) $-10, 10$

1. $$5x^{2}+4cx+c^{2}-20=0$$
2. $$400-4c^{2}=0$$
3. Substitute the line into the circle: $x^{2}$ + (2x + c)$^{2}$ = 20, giving $5x^{2}$ + 4cx + $c^{2}$ $-$ 20 = 0.
4. A tangent meets the circle at exactly one point, so this quadratic has one repeated root. Its discriminant is zero.
5. (4c)$^{2}$ $-$ 4 $\times$ 5($c^{2}$ $-$ 20) = 400 $-$ $4c^{2}$ = 0.
6. Hence $c^{2}$ = 100 and c = $-10$ or c = 10.

• P1 Establishing $5x^{2}+4cx+c^{2}-20=0$ or an equivalent valid method.
• P1 Establishing $400-4c^{2}=0$ or an equivalent valid method.
• A1 Correct answer: $-10, 10$