Question

5 the diagram below shows two points a(p,0) and b(p√2,0), and a circle centred at the point c(0,√7). the circle passes through a and intersects the line - segment bc at the point d. given that bd = 1 unit, find the value of p exactly.

Explanation:

Step1: Find the radius of the circle

The radius $r$ of the circle with center $C(0,\sqrt{7})$ and passing through $A(p,0)$ is given by the distance formula $r = \sqrt{(p - 0)^2+(0-\sqrt{7})^2}=\sqrt{p^{2}+7}$.

Step2: Find the distance between $B(p\sqrt{2},0)$ and $C(0,\sqrt{7})$

The distance $BC$ is calculated using the distance formula $d=\sqrt{(p\sqrt{2}-0)^2+(0 - \sqrt{7})^2}=\sqrt{2p^{2}+7}$.

Step3: Use the property of line - circle intersection

We know that $BC=BD + DC$. Since $DC$ is the radius $r=\sqrt{p^{2}+7}$ and $BD = 1$, we have $\sqrt{2p^{2}+7}=1+\sqrt{p^{2}+7}$.
Square both sides: $2p^{2}+7=(1 + \sqrt{p^{2}+7})^2=1 + 2\sqrt{p^{2}+7}+p^{2}+7$.
Simplify the equation:

$$
LATEXBLOCK0
$$

Square both sides again: $(p^{2}-1)^2 = 4(p^{2}+7)$.
Expand: $p^{4}-2p^{2}+1=4p^{2}+28$.
Rearrange to get a quartic equation: $p^{4}-6p^{2}-27 = 0$.
Let $x = p^{2}$, then $x^{2}-6x - 27=0$.
Factor the quadratic equation: $(x - 9)(x+3)=0$.
So $x = 9$ or $x=-3$. Since $x = p^{2}\geq0$, we take $x = p^{2}=9$, then $p = 3$ (we consider the positive value of $p$ as it represents a coordinate in a geometric context).

Answer:

$p = 3$