If the straight lines $x-4y+7=0$ and $3x-12y+11=0$ are tangents to a circle. Then, the radius of the circle is:
(A) $\dfrac{10}{3\sqrt{17}}$
(B) $\dfrac{5}{3\sqrt{7}}$
(C) $\dfrac{15}{\sqrt{17}}$
(D) $\dfrac{5}{3\sqrt{17}}$
(E) $\dfrac{5}{3\sqrt{13}}$
Answer
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Hint: For answering this question we will find the distance between the two given tangents as it is known as diameter for a circle and we know that the radius of a circle is half of the diameter of the circle. For finding the distance between any two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ we will use the formulae given as $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ .
Complete step by step answer:
Now considering from the question we have 2 straight lines which are the tangents of a circle. Then we can say that the distance between the 2 straight lines will be diameter because we know that the distance between 2 tangents is the diameter.
As we need the radius of the circle we will find the distance between the given 2 straight lines and divide it by 2 because the diameter is twice the radius.
The given straight lines are $x-4y+7=0$ and $3x-12y+11=0$ .
The distance between any two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given as $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.
Using the above formulae we can say that the distance between $x-4y+7=0$ and $x-4y+\dfrac{11}{3}=0$ which is similar to $3x-12y+11=0$ is given as $\dfrac{\left| 7-\dfrac{11}{3} \right|}{\sqrt{{{1}^{2}}+{{4}^{2}}}}$ .
Hence the diameter of the circle will be $\dfrac{10}{3\sqrt{17}}$ .
So the radius will be $\dfrac{5}{3\sqrt{17}}$ .
Hence we can conclude that for a circle having these 2 straight lines $x-4y+7=0$ and $3x-12y+11=0$ as tangents the radius will be $\dfrac{5}{3\sqrt{17}}$ .
Hence, option D is correct.
Note: While answering questions of this type we should make a note that the diameter is twice the radius and not the radius itself. And we should divide the distance between the two parallel lines by 2 otherwise we will obtain a wrong option that is A in this case.
Complete step by step answer:
Now considering from the question we have 2 straight lines which are the tangents of a circle. Then we can say that the distance between the 2 straight lines will be diameter because we know that the distance between 2 tangents is the diameter.
As we need the radius of the circle we will find the distance between the given 2 straight lines and divide it by 2 because the diameter is twice the radius.
The given straight lines are $x-4y+7=0$ and $3x-12y+11=0$ .
The distance between any two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given as $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.
Using the above formulae we can say that the distance between $x-4y+7=0$ and $x-4y+\dfrac{11}{3}=0$ which is similar to $3x-12y+11=0$ is given as $\dfrac{\left| 7-\dfrac{11}{3} \right|}{\sqrt{{{1}^{2}}+{{4}^{2}}}}$ .
Hence the diameter of the circle will be $\dfrac{10}{3\sqrt{17}}$ .
So the radius will be $\dfrac{5}{3\sqrt{17}}$ .
Hence we can conclude that for a circle having these 2 straight lines $x-4y+7=0$ and $3x-12y+11=0$ as tangents the radius will be $\dfrac{5}{3\sqrt{17}}$ .
Hence, option D is correct.
Note: While answering questions of this type we should make a note that the diameter is twice the radius and not the radius itself. And we should divide the distance between the two parallel lines by 2 otherwise we will obtain a wrong option that is A in this case.
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