
Sum of the squares of the lengths of the tangents from the points \[\left( {20,30} \right),\left( {30,40} \right),\left( {40,50} \right)\] to the circle \[{x^2} + {y^2} = 16\] is $a$. Find the value of \[\dfrac{{\left[ {\sqrt a } \right]}}{{11}}\], where [.] denotes greatest integer function.
(Consider only one tangent from each of the given points).
Answer
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Hint: Here we will find the squares of length of tangents from each of the points to the given circle by putting each point in the equation of the circle and then add them to find the value of a and then finally find the required value.
Complete step by step answer:
The given equation of the circle is:-
\[{x^2} + {y^2} = 16\]
Simplifying it we get:-
\[{x^2} + {y^2} - 16 = 0\]
Now we will find the square of length of the tangent from the point \[\left( {20,30} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {20} \right)^2} + {\left( {30} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 400 + 900 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 1300 - 16\]
\[\Rightarrow length = 1284\]……………….. (1)
Now we will find the square of length of the tangent from the point\[\left( {30,40} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {30} \right)^2} + {\left( {40} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 900 + 1600 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 2500 - 16\]
\[\Rightarrow length = 2484\]…………………… (2)
Now we will find the square of length of the tangent from the point\[\left( {40,50} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {40} \right)^2} + {\left( {50} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 1600 + 2500 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 4100 - 16\]
\[\Rightarrow length = 4084\]………………… (3)
Now adding the values from eq 1, 2 and 3 and then equate it equal to a we get:-
\[\Rightarrow 1284 + 2484 + 4084 = a\]
Simplifying it we get:-
\[\Rightarrow a = 7852\]
Now we need to find the value of\[\dfrac{{\left[ {\sqrt a } \right]}}{{11}}\].
Hence Putting the values we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{\left[ {\sqrt {7852} } \right]}}{{11}}\]
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{\left[ {88.6115} \right]}}{{11}}\]
Now since, [.] denotes greatest integer function.
Hence, we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{88}}{{11}}\]
Simplifying it we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = 8\]
Hence, the required value is 8.
Note:
Students should note that the greatest integer function of any value is less than or equal to the value i.e. it rounds the value inside the function to the nearest integer less than that value.
Complete step by step answer:
The given equation of the circle is:-
\[{x^2} + {y^2} = 16\]
Simplifying it we get:-
\[{x^2} + {y^2} - 16 = 0\]
Now we will find the square of length of the tangent from the point \[\left( {20,30} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {20} \right)^2} + {\left( {30} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 400 + 900 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 1300 - 16\]
\[\Rightarrow length = 1284\]……………….. (1)
Now we will find the square of length of the tangent from the point\[\left( {30,40} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {30} \right)^2} + {\left( {40} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 900 + 1600 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 2500 - 16\]
\[\Rightarrow length = 2484\]…………………… (2)
Now we will find the square of length of the tangent from the point\[\left( {40,50} \right)\].
Hence we need to substitute this point in the equation of the circle to get the square of the length of the tangent.
Putting in the values we get:-
\[ = {\left( {40} \right)^2} + {\left( {50} \right)^2} - 16\]
Solving it further we get:-
\[\Rightarrow length = 1600 + 2500 - 16\]
Simplifying it further we get:-
\[\Rightarrow length = 4100 - 16\]
\[\Rightarrow length = 4084\]………………… (3)
Now adding the values from eq 1, 2 and 3 and then equate it equal to a we get:-
\[\Rightarrow 1284 + 2484 + 4084 = a\]
Simplifying it we get:-
\[\Rightarrow a = 7852\]
Now we need to find the value of\[\dfrac{{\left[ {\sqrt a } \right]}}{{11}}\].
Hence Putting the values we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{\left[ {\sqrt {7852} } \right]}}{{11}}\]
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{\left[ {88.6115} \right]}}{{11}}\]
Now since, [.] denotes greatest integer function.
Hence, we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = \dfrac{{88}}{{11}}\]
Simplifying it we get:-
\[\Rightarrow \dfrac{{\left[ {\sqrt a } \right]}}{{11}} = 8\]
Hence, the required value is 8.
Note:
Students should note that the greatest integer function of any value is less than or equal to the value i.e. it rounds the value inside the function to the nearest integer less than that value.
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