Question

Find the shortest distance from the points M (− 7, 2) to the circle ${x^2} + {y^2} - 10x - 14y - 151 = 0$

Answer 1

Hint: We have the general equation of the circle is, $a{x^2} + b{y^2} + 2gx + 2gy + c = 0$ then comparing the given equation by general equation of the circle, find radius of the circle and centre of the circle. Then find the length of line from the centre to the point M and also find the length of radius, the difference between the two lengths will be the shortest distance from the points as given below.

Complete step by step answer:
Let the equation of the circle be, $S = {x^2} + {y^2} - 10x - 14y - 151 = 0$
Now putting the value of points (−7,2) in the equation to check whether the given point is on the circle, inside the circle or outside the circle.
$S = - {7^2} + {2^2} - 10 \times ( - 7) - 14 \times (2) - 151 = 0$
\[S = \;49 + 4 + 70 - 28 - 151 = - 56\] \[\]
S= -56 <0 \[S = - 56 < 0\]
Therefore, M (-7,2) does not lie on the circle.
The general equation of the circle is, $a{x^2} + b{y^2} + 2gx + 2gy + c = 0$
Now, g = 5, f = 7, c = −151
The radius of the circle is,
$r = \sqrt {{g^2} + {f^2} - c} $
$ = \sqrt {({5^2} + ({7^2}) - ( - 151)} $
\[ = \sqrt {25 + 49 + 151} = \sqrt {225} = 15\]
The centre of the circle is, C = (− g , − f) = ( 5,7)
Now, $CM = \sqrt {{{(5 + 7)}^2} + {{(7 - 2)}^2}} $
CM= 122+(52)
$ \Rightarrow CM = \sqrt {{{12}^2} + {5^2}} $
$ \Rightarrow CM = \sqrt {144 + 25} $
$ \Rightarrow CM = \sqrt {169} = 13$
The shortest distance is, r – CM = 15 – 13 = 2
Hence the shortest distance is 2 units.

Note: In this type of question, we must know about the general equation for substitute the points in equation which is already given in question and you must know about the radius of the circle then put the value g, f and c for finding radius with carefully after that, put the value for finding centre of the circle and you must calculate of roots with sharply then you can find the shortest distance of difference r and CM.