
The distance between the chords of contact of tangents to the circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]from the origin and from the point (g, f) is
( A ) \[\sqrt {{g^2} + {f^2}} \]
( B ) \[\dfrac{{\sqrt {{g^2} + {f^2} - c} }}{2}\]
( C ) \[\dfrac{{{g^2} + {f^2} - c}}{{2\sqrt {{g^2} + {f^2}} }}\]
( D ) \[\dfrac{{\sqrt {{g^2} + {f^2} + c} }}{{2\sqrt {{g^2} + {f^2}} }}\]
Answer
582.6k+ views
Hint: Find the equation of both the chord of contact then use the distances between lines formula.
Equation of chord of contact from (p, q) is \[xp + yq + g(x + p) + f(y + q) + c = 0\]
two lines are given as:
\[ax + by + {c_1} = 0\]and \[{a_{}}x + {b_{}}y + {c_2} = 0\]
Distance between two parallel line is: \[\begin{array}{l}
D = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|\\
\end{array}\]
Given: given equation of circle is: - \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
Complete step-by-step answer:
When we are solving this type of questions, we need to follow the steps
provided in the hint part above.
equations of chord of contact of tangents from origin (0,0) and point (g, f) given circle are,
\[\begin{array}{l}
x(0) + y(0) + g(x + 0) + f(y + 0) + c = 0\\
\Rightarrow gx + fy + c = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\\
xg + yf + g(x + g) + f(y + f) + c = 0\\
\Rightarrow gx + fy + \dfrac{{({g^2} + {f^2} + c)}}{2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)
\end{array}\]
From (1) and (2) it's clear that equations are parallel.
\[D = \left| {\dfrac{{\dfrac{{({g^2} + {f^2} + c)}}{2} - c}}{{\sqrt {{g^2} + {f^2}} }}} \right|\]
\[D = \left| {\dfrac{{{g^2} + {f^2} - c}}{{2\sqrt {{g^2} + {f^2}} }}} \right|\]
Hence option (D) is correct
Hence after following the each and every step given in the hint part we obtained our final answer.
Additional Information: Here we can clearly see that in this solution we did not use any complicated process because we followed basic and simple things in the right order as per given in the above hint section.
Note: In this sort of example we need to face the various things and some of them are referred to here which will be really helpful to fathom the main concept of the problem.
We need to use the right formula to avoid unnecessary things which are sometimes time consuming.
Equation of chord of contact from (p, q) is \[xp + yq + g(x + p) + f(y + q) + c = 0\]
two lines are given as:
\[ax + by + {c_1} = 0\]and \[{a_{}}x + {b_{}}y + {c_2} = 0\]
Distance between two parallel line is: \[\begin{array}{l}
D = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|\\
\end{array}\]
Given: given equation of circle is: - \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
Complete step-by-step answer:
When we are solving this type of questions, we need to follow the steps
provided in the hint part above.
equations of chord of contact of tangents from origin (0,0) and point (g, f) given circle are,
\[\begin{array}{l}
x(0) + y(0) + g(x + 0) + f(y + 0) + c = 0\\
\Rightarrow gx + fy + c = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\\
xg + yf + g(x + g) + f(y + f) + c = 0\\
\Rightarrow gx + fy + \dfrac{{({g^2} + {f^2} + c)}}{2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)
\end{array}\]
From (1) and (2) it's clear that equations are parallel.
\[D = \left| {\dfrac{{\dfrac{{({g^2} + {f^2} + c)}}{2} - c}}{{\sqrt {{g^2} + {f^2}} }}} \right|\]
\[D = \left| {\dfrac{{{g^2} + {f^2} - c}}{{2\sqrt {{g^2} + {f^2}} }}} \right|\]
Hence option (D) is correct
Hence after following the each and every step given in the hint part we obtained our final answer.
Additional Information: Here we can clearly see that in this solution we did not use any complicated process because we followed basic and simple things in the right order as per given in the above hint section.
Note: In this sort of example we need to face the various things and some of them are referred to here which will be really helpful to fathom the main concept of the problem.
We need to use the right formula to avoid unnecessary things which are sometimes time consuming.
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