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This is one of our many quizzes on Tangent lines. Click on this to access more such competitive quizzes on Geometry.

Tangent lines Quiz 2

mathematics

 The tangents ABC and DEF touch the three circles with centres ${{O}_{1}},\text{ }{{O}_{2}}\text{ and }{{O}_{3}}$ and radii 3 cm, 4 cm and 12 cm respectively as shown in the figure below. If $AC=39cm$ and $EF=15cm$, then what is the distance between ${{O}_{1}}\text{ and }{{O}_{2}},\text{ }{{O}_{2}}\text{ and }{{O}_{3}} \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e419ffe7-8192-4f18-823e-d5948dbe52681989152586314185304.png"/>

  Find the lengths by using $AB=\sqrt{{{(\text{distance between }{{O}_{1}}\text{ and }{{O}_{2}})}^{2}}-{{({{r}_{1}}+{{r}_{2}})}^{2}}}$and $BC=\sqrt{{{(\text{distance between }{{O}_{2}}\text{ and }{{O}_{3}})}^{2}}-{{({{r}_{3}}+{{r}_{2}})}^{2}}}$ formula.

$\text{Given, }{{r}_{1}}=3cm,{{r}_{2}}=4cm\text{ and }{{r}_{3}}=12cm \\ \text{AB and DE are the transverse common tangents } \text{while BC and EF are the direct common tangents.} \\ \text{Now, } AB=DE\text{ and }BC=EF \\ \Rightarrow BC=15cm\text{ and }AB=39-15=24cm \\ AB=\sqrt{{{(\text{distance between }{{O}_{1}}\text{ and }{{O}_{2}})}^{2}}-{{({{r}_{1}}+{{r}_{2}})}^{2}}} \\ 24=\sqrt{{{(\text{distance between }{{O}_{1}}\text{ and }{{O}_{2}})}^{2}}-{{(3+4)}^{2}}} \\ \Rightarrow {{(\text{distance between }{{O}_{1}}\text{ and }{{O}_{2}})}^{2}}=625 \\ \Rightarrow \text{distance between }{{O}_{1}}\text{ and }{{O}_{2}}=25 \\ \text{And } BC=\sqrt{{{(\text{distance between }{{O}_{2}}\text{ and }{{O}_{3}})}^{2}}-{{({{r}_{3}}+{{r}_{2}})}^{2}}} \\ 15=\sqrt{{{(\text{distance between }{{O}_{2}}\text{ and }{{O}_{3}})}^{2}}-{{(12-4)}^{2}}} \\ \Rightarrow (\text{distance between }{{O}_{2}}\text{ and }{{O}_{3}})^2=289 \\ \Rightarrow \text{distance between }{{O}_{2}}\text{ and }{{O}_{3}}=17$

 The 10 cm line PQ makes an angle of ${{45}^{0}}$ with the tangent line PT. If O is the centre of the circle given below then find the approximate perimeter of the $\Delta MOP \\ $.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c37bbedf-27bd-42df-9647-aedbc49be2c63153811514590688358.png"/>

  First find $\angle MOP$ , then the length of $OP$, after that find the length of $MP$ by using cosine formula to find the perimeter.


$\angle OPT={{90}^{\circ }},\angle QPT={{45}^{\circ }} \\ \angle OPQ={{45}^{\circ }},\angle PON={{45}^{\circ }} \\ \angle MOP={{135}^{\circ }} \\ \text{Now, }PQ=10cm \Rightarrow PN=5cm=NO  \\ OP=5\sqrt{2}cm=OM \\ \text{Again, } \\ M{{P}^{2}}=O{{M}^{2}}+O{{P}^{2}}-2OM\times OP\times \cos {{135}^{0}} \\ 2{{(OM)}^{2}}-2{{(OM)}^{2}}\cos {{135}^{0}}=2{{(5\sqrt{2})}^{2}}-2{{(5\sqrt{2})}^{2}}\times \left( -\dfrac{1}{\sqrt{2}} \right) \\ MP\approx 13cm \\ \text{Perimeter of }\Delta MOP=5\sqrt{2}+5\sqrt{2}+13=27cm$

 In the figure, 'O' is the centre of circle and a tangent at point 'C', then $\angle q$ is: $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e73d51db-a37c-4587-bbc7-0afbe0a03cde6945280319123567078.png"/>

  By the formula, Angle formed from the diameter to the circumference of the circle is always a right angle find $\angle q$.

$\text{From the fig, }\angle ACB={{90}^{0}}~~~~~~~~ \text{[Angle formed from the diameter} \text{ to the circumference of the circle]} \\ \angle CAB={{180}^{0}}-\left( {{90}^{0}}+{{30}^{0}} \right)={{60}^{0}}\text{  [From Alternate segment theorem] } \\ \text{Therefore, }\angle q={{60}^{0}}$

 In the given figure, if the radius of the circles with centres P and Q are 256 cm and 16 cm respectively and MN is the common tangent drawn to both the circles, then what is the radius of the third circle with centre A which touches the circles with centres P and Q and the line MN externally as shown in the figure? $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/62e9a656-0322-40d5-af2c-7130a43513255610511434978825520.png"/>

  Since we know that, if two circles of radii R and r are externally tangent, the length of the direct common tangent $=\sqrt{4Rr}$, we will find out PN, QM and OA.

$\text{If two circles of radii R and r are externally tangent, } \text{the length of the direct common tangent }=\sqrt{4Rr} \\ \text{Let the points at which the line is tangent} \text{ to the circles with centres }P, Q, O\text{ be }M, N\text{ and }A\text{ respectively.} \\ \text{Let } PN=a,QM=b\text{ and OA=c} \\ AN \text{=}\sqrt{4ac},MA=\sqrt{4bc}\text{ and }MN=\sqrt{4ab} \\ \therefore \sqrt{4ac}+\sqrt{4bc}=\sqrt{4ab} \\ \Rightarrow \sqrt{c}=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{256\times 16}}{\sqrt{256}+\sqrt{16}} \\ \Rightarrow c=10.24cm$

 Two circles are shown below, $NO$ is their common tangent, they intersect at two points $M\text{ and }P$ such that $MN=MO\text{ and }NP=OP$. If $\angle MNP={{40}^{0}}$, then find $\angle NOP \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ed75be51-f6e1-444e-a222-611399af2f1c247394565460017233.png"/>

  First, find $\angle MOP$, then by the theorem, the angle between a tangent and a chord is equal to the angle in the alternate segment, we will get the other two angles, from there if we consider $\Delta MNO$, we will get $\angle NOP$.

$\text{We have, }MN=MO, \angle MNO=\angle MON \\ \text{And }NP=OP,\angle PNO=\angle NOP \\ \text{Therefore, }\angle MNO-\angle PNO=\angle MNP\text{ and }\angle MON-\angle PON=\angle MOP \\ \text{Thus, }\angle MOP={{40}^{0}} \\ \text{As, the angle between a tangent and a chord is equal } \text{to the angle in the alternate segment, we have,} \\ \angle NMP=\angle PNO\text{ and }\angle OMP=\angle PON \\ \text{Let }\angle PNO=\angle NOP={{x}^{0}}\text{ then }\angle NMP=\angle OMP={{x}^{0}} \\ \text{In }\Delta MNO \\ {{40}^{0}}+{{x}^{0}}+{{40}^{0}}+{{x}^{0}}+{{x}^{0}}+{{x}^{0}}={{180}^{0}} \\ x={{25}^{0}}$