Question

$P$ is a point of intersection of two circles with centres $C$ and $D$. If the straight line $APB$ is parallel to $CD$, then

• A. $AB=CD$
• B. $AB=\frac{1}{2}CD$
• C. $3AB=CD$
• D. $AB=2CD$

Answer 1

Answer: (D)

$AB=2CD$

$\because AB$ is parallel to $CD$

$\Rightarrow C{C}^1=D{D}^1$

and hence, $C^1{d}^1=CD\to \left(a\right)$

In $△A{C}^{1}C$ and PC^{1}C$

$\Rightarrow AC=PC$=radius of $I$ circle.

$\angle C{C}^{1}A=\angle C{C}^{1}P={90}^0$

$C{C}^1=C{C}^1$ (common)

$\therefore A{C}^{1}C\cong P{C}^{1}C$ by R.H.S.

$\Rightarrow A{C}^1=P{C}^1\to \left(1\right)$

Similarly, $P{D}^1=B{D}^1\to \left(2\right)$

$\therefore AB=A{C}^1+P{C}^1+P{D}^1+B{D}^1$

Using $\left(1\right)$ and $\left(2\right)$

$\Rightarrow AB=2P{C}^1+2P{D}^1$

$\Rightarrow AB=2(P{C}^1+P{D}^1)$

$\Rightarrow AB=2C^1{D}^1$

From $\left(a\right)$ $AB=2CD$

$\therefore D)$ Answer.