Question

Two circles with centres $ \mathrm{O}$ and $ {\mathrm{O}}^{\text{'}}$ of radii $ 3 \mathrm{cm}$ and $ 4 \mathrm{cm}$, respectively intersect at two points $ \mathrm{P}$ and $ \mathrm{Q}$ such that $ \mathrm{OP}$ and $ {\mathrm{O}}^{\text{'}}\mathrm{P}$ are tangents to the two circles. Find the length of the common chord $ \mathrm{PQ}$.

Answer 1

Radius:

The distance between the center and circumference of the circle is called a radius.

Tangent:

A line intersecting at only one external point of a circle is called a tangent.

Chord of the circle:

A line segment whose endpoints lie on a circle is called a chord of the circle.

Calculating the length of the common chord $\mathbf{PQ}$.

The center of the two circles is $\mathrm{O}$ and ${\mathrm{O}}^{\text{'}}$ with radius $\mathrm{OP}=3\mathrm{cm}$ and $\mathrm{O}\text{'}\mathrm{P}=4\mathrm{cm}$ respectively.

The two circles intersect at the points $\mathrm{P}$ and $\mathrm{Q}$.

The tangent of two circles is $\mathrm{OP}$ and ${\mathrm{O}}^{\text{'}}\mathrm{P}$.

The tangent at any point on the circle is perpendicular to the radius through the point of contact.

$\Rightarrow \angle {\mathrm{OPO}}^{\text{'}}=90°$

Thus, $△{\mathrm{OPO}}^{\text{'}}$ is a right-angled triangle.

Applying Pythagoras in $\mathbf{△}{\mathbf{OPO}}^{\mathbf{\text{'}}}$:

$$\begin{gathered} {\left({\mathrm{OO}}^{\text{'}}\right)}^2={\left(\mathrm{OP}\right)}^2+{\left({\mathrm{O}}^{\text{'}}\mathrm{P}\right)}^2 \\ \Rightarrow {\left({\mathrm{OO}}^{\text{'}}\right)}^2=3^2+4^2 \\ \Rightarrow {\left({\mathrm{OO}}^{\text{'}}\right)}^2=9+16 \\ \Rightarrow {\left({\mathrm{OO}}^{\text{'}}\right)}^2=25 \\ \Rightarrow {\mathrm{OO}}^{\text{'}}={\left(25\right)}^2 \\ \Rightarrow {\mathrm{OO}}^{\text{'}}=5\mathrm{cm} \end{gathered}$$

Consider x to be the length of ON.

Considering $\mathrm{ON}=\mathrm{x}\mathrm{cm}$.

$$\begin{gathered} \mathrm{NO}\text{'}=\mathrm{OO}\text{'}-\mathrm{ON} \\ \Rightarrow \mathrm{NO}\text{'}=5-\mathrm{x} \end{gathered}$$

Thus, $\mathrm{NO}\text{'}=(5-\mathrm{x})\mathrm{cm}$.

Applying Pythagoras in $\mathbf{△}\mathbf{OPN}$:

$$\begin{gathered} {\left(\mathrm{OP}\right)}^2={\left(\mathrm{ON}\right)}^2+{\left(\mathrm{PN}\right)}^2 \\ \Rightarrow 3^2={\mathrm{x}}^2+{\left(\mathrm{PN}\right)}^2 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=3^2-{\mathrm{x}}^2 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=9-{\mathrm{x}}^2\to \left(1\right) \end{gathered}$$

Applying Pythagoras in $\mathbf{△}\mathbf{O}\mathbf{\text{'}}\mathbf{PN}$:

$$\begin{gathered} {\left(\mathrm{O}\text{'}\mathrm{P}\right)}^2={(\mathrm{NO}\text{'})}^2+{\left(\mathrm{PN}\right)}^2 \\ \Rightarrow 4^2={(5-\mathrm{x})}^2+{\left(\mathrm{PN}\right)}^2 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=4^2{-(5-\mathrm{x})}^2 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=16-\left(5^2+{\mathrm{x}}^2-2\left(5\right)\mathrm{x}\right)\left[\mathrm{By}\mathrm{using}{(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\right] \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=16-25-{\mathrm{x}}^2+10\mathrm{x} \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=-{\mathrm{x}}^2+10\mathrm{x}-9\to \left(2\right) \end{gathered}$$

Equating equations $\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ and $\mathbf{\left(}\mathbf{2}\mathbf{\right)}$:

$$\begin{gathered} 9-{\mathrm{x}}^2=-{\mathrm{x}}^2+10\mathrm{x}-9 \\ \Rightarrow 9-{\mathrm{x}}^2-\left(-{\mathrm{x}}^2+10\mathrm{x}-9\right)=0 \\ \Rightarrow 9-{\mathrm{x}}^2+{\mathrm{x}}^2-10\mathrm{x}+9=0 \\ \Rightarrow -10\mathrm{x}+9+9=0 \\ \Rightarrow -10\mathrm{x}+18=0 \\ \Rightarrow -10\mathrm{x}=-18 \\ \Rightarrow 10\mathrm{x}=18 \\ \Rightarrow \mathrm{x}=\frac{18}{10} \\ \Rightarrow \mathrm{x}=1.8 \end{gathered}$$

Thus, $\mathrm{ON}=1.8\mathrm{cm}$.

Substituting the value of x in equation $\mathbf{\left(}\mathbf{1}\mathbf{\right)}$:

$$\begin{gathered} {\left(\mathrm{PN}\right)}^2=9-1.8^2 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=9-3.24 \\ \Rightarrow {\left(\mathrm{PN}\right)}^2=5.76 \\ \Rightarrow \mathrm{PN}=\pm \sqrt{5.76} \\ \Rightarrow \mathrm{PN}=\pm \sqrt{2.4\times 2.4} \\ \Rightarrow \mathrm{PN}=\pm 2.4 \end{gathered}$$

The length cannot be in negative.

Thus, the value of $\mathrm{PN}=2.4\mathrm{cm}$.

From the figure, the length of a chord $\mathrm{PQ}$ is two times the value of $\mathrm{PN}$.

$$\begin{gathered} \Rightarrow \mathrm{PQ}=2\left(\mathrm{PN}\right) \\ \Rightarrow \mathrm{PQ}=2(2.4) \\ \Rightarrow \mathrm{PQ}=4.8 \end{gathered}$$

Thus, the value of $\mathrm{PQ}=4.8\mathrm{cm}$.

Hence, the length of the chord of the two circles is $\mathbf{4}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{cm}$.