Circle 5 Question 1

1. If the angle of intersection at a point where the two circles with radii $5 cm$ and $12 cm$ intersect is $90^{\circ}$, then the length (in cm) of their common chord is

(a) $\frac{13}{5}$

(b) $\frac{120}{13}$

(c) $\frac{60}{13}$

(d) $\frac{13}{2}$

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Answer:

Correct Answer: 1. (b)

Solution:

  1. Let the length of common chord $=A B=2 A M=2 x$

Now, $C _1 C _2=\sqrt{A C _1^{2}+A C _2^{2}}$

$\left[\because\right.$ circles intersect each other at $\left.90^{\circ}\right]$

$ \begin{aligned} & \text { and } C _1 C _2=C _1 M+M C _2 \\ & \Rightarrow \quad C _1 C _2=\sqrt{12^{2}-A M^{2}}+\sqrt{5^{2}-A M^{2}} \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & \sqrt{A C _1^{2}+A C _2^{2}}=\sqrt{144-A M^{2}}+\sqrt{25-A M^{2}} \\ & \Rightarrow \quad \sqrt{144+25}=\sqrt{144-x^{2}}+\sqrt{25-x^{2}} \\ & \Rightarrow \quad 13=\sqrt{144-x^{2}}+\sqrt{25-x^{2}} \end{aligned} $

On squaring both sides, we get

$ \begin{aligned} & 169=144-x^{2}+25-x^{2}+2 \sqrt{144-x^{2}} \sqrt{25-x^{2}} \\ & \Rightarrow \quad x^{2}=\sqrt{144-x^{2}} \sqrt{25-x^{2}} \end{aligned} $

Again, on squaring both sides, we get

$x^{4}=\left(144-x^{2}\right)\left(25-x^{2}\right)=(144 \times 25)-(25+144) x^{2}+x^{4}$

$\Rightarrow x^{2}=\frac{144 \times 25}{169} \Rightarrow x=\frac{12 \times 5}{13}=\frac{60}{13} cm$

Now, length of common chord $2 x=\frac{120}{13} cm$

Alternate Solution

Given, $A C _1=12 cm$ and $A C _2=5 cm$

In $\Delta C _1 A C _2$,

$ C _1 C _2=\sqrt{\left(C _1 A\right)^{2}+\left(A C _2\right)^{2}} \quad\left[\because \angle C _1 A C _2=90^{\circ},\right. $

because circles intersects each other at $90^{\circ}$ ]

$ =\sqrt{169}=13 cm $

$ =\sqrt{(12)^{2}+(5)^{2}}=\sqrt{144+25} $

Now, area of $\Delta C _1 A C _2=\frac{1}{2} A C _1 \times A C _2$

$ =\frac{1}{2} \times 12 \times 5=30 cm^{2} $

Also, area of $\Delta C _1 A C _2=\frac{1}{2} C _1 C _2 \times A M$

$ =\frac{1}{2} \times 13 \times \frac{A B}{2} \quad \because A M=\frac{A B}{2} $

$ \begin{aligned} \therefore \quad \frac{1}{4} \times 13 \times A M & =30 cm \\ A M & =\frac{120}{13} cm \end{aligned} $



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