Maths M.C.Q

$A D$ and $A C$ are radii of same circle and $C D$ is a chord.
Consider $\triangle \mathrm{ABC}$,
$B C^2=(A C)^2-(A B)^2$
$=5^2-4^2=25-16=9$
$\Rightarrow B C=3 \mathrm{~cm}$
Chord $C D=2 \times B C=6 \mathrm{~cm}$

Both the circles pass through the centre of each other
$\Rightarrow \mathrm{O}_1 \mathrm{O}_2=\mathrm{r}$
Common chord is $AB$
We know that perpendicular drawn from centre of circle to any chord bisects it.
$\Rightarrow P$ is the midpoint of $A B$
$\Rightarrow \mathrm{PA}=\mathrm{PB}$
$\mathrm{O}_1 \mathrm{~A}=\mathrm{r}$ (radius of circle)
Consider $\triangle\text{O}_1\text{PA}$
$\big(\text{O}_1\text{A}\big)^2=\text{AP}^2+\text{O}_1\text{P}^2$
$\Rightarrow\text{r}^2=\text{AP}^2+\Big(\frac{\text{r}}{2}\Big)^2$ ...($P$ is also mid-point of $\mathrm{O_1O_2}$)
$\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
$\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$

$A B=8 \mathrm{~cm}$
$\Rightarrow A C=B C=4 \mathrm{~cm}$
Consider $\triangle \mathrm{OCB}$, where $\mathrm{BC}=8 \mathrm{~cm}$,
$O B=6 \mathrm{~cm}$
$\text { Now, }(O C)^2+(B C)^2=(O B)^2$
$\Rightarrow(O C)^2+4^2=6^2$
$\Rightarrow(O C)^2+16=36$
$\Rightarrow(O C)^2=20$
$\Rightarrow O C=\sqrt{20}=2 \sqrt{5}$

$\mathrm{PQ}=23 \mathrm{~cm}$
$\mathrm{AB}=16 \mathrm{~cm}$
$\Rightarrow B P=A P=8 \mathrm{~cm}$
$\mathrm{r}=17 \mathrm{~cm}$
$\Rightarrow E F=\text { diameter }=2 \mathrm{r}=34 \mathrm{~cm}$
Consider $\triangle \mathrm{OPB}$,
$r^2=O P^2+B P^2$
$\Rightarrow O P^2=(17)^2-(8)^2=289-64=225$
$\Rightarrow O P=15 \mathrm{~cm}$
$\Rightarrow O Q=23-15=8 \mathrm{~cm}$
Consider $\triangle \mathrm{OQD}$,
$r^2=O Q^2+Q D^2$
$\Rightarrow Q D^2=r^2-O Q^2=(17)^2-(8)^2=225$
$\Rightarrow O D=15 \mathrm{~cm}$
$\Rightarrow C D=2 \times Q D=30 \mathrm{~cm}$