# CIE A level -Pure Mathematics 1 : Topic : 1.3 Coordinate geometry: solve problems involving lines and circles : Exam Questions Paper 1

### Question

The diagram shows the curve with equation $$y = 4x^{\frac{1}{2}}$$

(i) The straight line with equation $$y = x + 3$$ intersects the curve at points A and B. Find the length
of AB.

(ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T.

(iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB.

(i) Length of AB:
We have the equation $$4x^{1/2} = x + 3$$. Let’s solve this equation:
$$16x = (x + 3)^2$$
$$16x = x^2 + 6x + 9$$
$$x^2 – 10x +9 = 0$$
$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
For our equation, $$a = 1$$, $$b = -10$$, and $$c = 9$$.
$$x = \frac{10 \pm \sqrt{(-10)^2 – 4(1)(9)}}{2(1)}$$
$$x = \frac{10 \pm \sqrt{100-36}}{2}$$
$$x = \frac{10 \pm \sqrt{64}}{2}$$
$$x = \frac{10 \pm 2\sqrt{64}}{2}$$
$$x = \frac{10\pm 8}{2}$$
$$\x=frac{10+8}{2}$$  and $$x= \frac{10-8}{2}$$
x=9 and x=1
So, we have two values for $$x$$:9 and 1
Now, substitute these values into either the equation of the curve or the line to find the corresponding $$y$$-coordinates. Let’s use the equation of the curve:
$$y = 4x^{1/2}$$
For x=9
$$y = 4(9)^{1/2}$$
y=12
Similarly, for x=1
$$y = 4(1)^{1/2}$$
y=4
Now, we have the coordinates of points A and B. Use the distance formula to find the length of $$AB$$:
$$\text{Length of } AB = \sqrt{(x_A – x_B)^2 + (y_A – y_B)^2}$$
$$AB=\sqrt{(9-1)^{2}+(12-4)^{2}}$$
$$AB=\sqrt{8^{2}+8^{2}}$$
$$AB=\sqrt{64+64}$$
$$AB=\sqrt{128}$$
$$AB=11.3$$
(ii) The tangent to the curve at point T is parallel to AB. For a parallel tangent, the slopes of the curve and the line are equal. Find the slope of the line $$y = x + 3$$ (which is 1), and set it equal to the derivative of the curve’s equation with respect to $$x$$:
$$\frac{dy}{dx} = \frac{d}{dx}(4x^{1/2})$$
$$\frac{1}{2} \cdot 4x^{-1/2} = 1$$
$$2x^{-1/2} = 1$$
$$x=4$$
Now substitute x into the curve equation:
$$y=4x^{\frac{1}{2}}$$
$$y=8$$
The coordinates $$(4,8)$$ are the coordinates of point T.
(iii)The equation of the normal is given by:
$$y – 8 = -1(x – 4)$$
Now, let’s eliminate $$y$$ or $$x$$:
$$-x + 12 = x + 3$$
$$2x = 9$$
$$x = \frac{9}{2}$$
Now, substitute $$x$$ back into the equation of the normal to find $$y$$:
$$y – 8 = -1\left(\frac{9}{2} – 4\right)$$
$$y – 8 = -1\left(\frac{1}{2}\right)$$
$$y – 8 = -\frac{1}{2}$$
$$y = \frac{15}{2}$$
So, the point of intersection is $$\left(\frac{9}{2}, \frac{15}{2}\right)$$, which is equivalent to $$\left(4^{\frac{1}{2}}, 7^{\frac{1}{2}}\right)$$.

### Question

A diameter of a circle $$C_1$$ has end-points at (-3,-5) and (7,3).
(a) Find an equation of the circle $$C_1$$.

The circle $$C_1$$ is translated by $$\binom{8}{4}$$ to give circle $$C_2$$, as shown in the diagram.
(b) Find an equation of the circle $$C_2$$.
The two circles intersect at points R and S.
(c) Show that the equation of the line RS is y=-2x+13.
(d) Hence show that the x-coordinates of R and S satisfy the equation $$5x^2-60x+159=0$$.

(a) Equation of Circle $$C_1$$:
The equation of a circle with center $$(h, k)$$ and radius $$r$$ is given by $$(x – h)^2 + (y – k)^2 = r^2$$.
Given the endpoints of the diameter $$(-3, -5)$$ and $$(7, 3)$$, the center $$(h, k)$$ is the midpoint of the diameter, which can be found as follows:
$$h = \frac{-3 + 7}{2} = 2$$
$$k = \frac{-5 + 3}{2} = -1$$
The radius $$r$$ is the distance from the center to one of the endpoints. Using the distance formula:
$$r = \sqrt{(7 – 2)^2 + (3 – (-1))^2} = \sqrt{25 + 16} = \sqrt{41}$$
Therefore, the equation of circle $$C_1$$ is:
$$(x – 2)^2 + (y + 1)^2 = 41$$
(b) Equation of Circle $$C_2$$:
The circle $$C_2$$ is obtained by translating $$C_1$$ by $$\begin{bmatrix} 8 \\ 4 \end{bmatrix}$$. The center of $$C_2$$ is the sum of the center of $$C_1$$ and the translation vector:
$$\text{Center of } C_2 = \begin{bmatrix} 2 + 8 \\ -1 + 4 \end{bmatrix} = \begin{bmatrix} 10 \\ 3 \end{bmatrix}$$
The radius of $$C_2$$ remains the same as $$C_1$$, which is $$\sqrt{41}$$. Therefore, the equation of $$C_2$$ is:
$$(x – 10)^2 + (y – 3)^2 = 41$$
(c)$$(x-2)^2+(y+1)^2-41=(x-10)^2+(y-3)^2-41OE$$
$$x^2-4x+4+y^2+2y+1=x^2-20x+100+y^2-6y+9 OE$$
16x+8y=104
y=-2x+13
(d) $$(x-10)^2(-2x+13-3)^2=41$$
$$x^2-20x+100+4x^2-40x+100=41 \rightarrow 5x^2-60x+159=0$$

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