A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is
A. m(m + n)
B. m + n
C. n(m + n)
D. \[\dfrac{1}{2}\left( {m + n} \right)\]
Answer
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Hint: Here, we first find the centre of the circle and obtaining this we can easily find the radius of the circle. Using the equation of tangent passing through the circle, we can find the lengths m and n. As we know that diameter of a circle is twice its radius, so using this we can calculate diameter.
Complete step-by-step answer:
Given coordinates of A is (a, 0) and that of B is (0, b).
Since ∠AOB = $\dfrac{\pi }{2}$, as given that the line AB is a diameter of the circle circumscribing the triangle OAB, its centre is the mid-point of AB, i.e., \[\left( {\dfrac{a}{2},\dfrac{b}{2}} \right)\], and its radius is \[\dfrac{1}{2}AB = \dfrac{1}{2}\sqrt {{a^2} + {b^2}} \] . Therefore, the equation of the circle through O, A and B is \[{x^2} + {y^2} - ax - by = 0\] and the equation of the tangent at the origin to this circle is ax + by = 0.
If AL and BM are the perpendiculars from A and B to this tangent, then
\[AL = m = \dfrac{{{a^2}}}{{\sqrt {{a^2} + {b^2}} }}\] and \[BM = n = \dfrac{{{b^2}}}{{\sqrt {{a^2} + {b^2}} }}\]
Hence, m + n\[ = \sqrt {{a^2} + {b^2}} \] is the diameter of the circle.
So, the correct answer is “Option B”.
Note: In these types of questions, first estimation the lengths of different lengths using properties. As in coordinate geometry, coordinates itself associated with length calculate length whenever find coordinates of the endpoint of line. Careful about the equation of the circle, because the standard equation is only when the centre is origin. Tangent to the circle is always perpendicular at the point of contact. Always remember the radius of a circle is half the diameter of the circle. In coordinate geometry, we can say that the centre of a circle is the midpoint of endpoints of diameter of circle.
Complete step-by-step answer:
Given coordinates of A is (a, 0) and that of B is (0, b).
Since ∠AOB = $\dfrac{\pi }{2}$, as given that the line AB is a diameter of the circle circumscribing the triangle OAB, its centre is the mid-point of AB, i.e., \[\left( {\dfrac{a}{2},\dfrac{b}{2}} \right)\], and its radius is \[\dfrac{1}{2}AB = \dfrac{1}{2}\sqrt {{a^2} + {b^2}} \] . Therefore, the equation of the circle through O, A and B is \[{x^2} + {y^2} - ax - by = 0\] and the equation of the tangent at the origin to this circle is ax + by = 0.
If AL and BM are the perpendiculars from A and B to this tangent, then
\[AL = m = \dfrac{{{a^2}}}{{\sqrt {{a^2} + {b^2}} }}\] and \[BM = n = \dfrac{{{b^2}}}{{\sqrt {{a^2} + {b^2}} }}\]
Hence, m + n\[ = \sqrt {{a^2} + {b^2}} \] is the diameter of the circle.
So, the correct answer is “Option B”.
Note: In these types of questions, first estimation the lengths of different lengths using properties. As in coordinate geometry, coordinates itself associated with length calculate length whenever find coordinates of the endpoint of line. Careful about the equation of the circle, because the standard equation is only when the centre is origin. Tangent to the circle is always perpendicular at the point of contact. Always remember the radius of a circle is half the diameter of the circle. In coordinate geometry, we can say that the centre of a circle is the midpoint of endpoints of diameter of circle.
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