logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the shortest and largest distance from the point (2,-7) to the circle $x^2+y^2-14x-10y-151=0$.

$\begin{array}{1 1}(a)\;2\;and\;28\\(b)\;4\;and\;20\\(c)\;3\;and\;27\\(d)\;5\;and\;25\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Let $S=x^2+y^2-14x-10y-151=0$
Substituting the point $(2,-7)$ in this equation, we get
$S_1=(2)^2+(-7)^2-14(2)-10(-7)-151=-56 < 0$
$\therefore P(2,-7)$ inside the circle
Radius of circle $r=\sqrt{(-7)^2+(-5)^2+151}=15$
Centre of circle (7,5)
CP=$\sqrt{(7-2)+(5+7)^2}=13$
Shortest distance =$PA=r-CP=15-13=2$
Largest distance $PB=r+CP=15+13=28$
Hence (a) is the correct option.
answered Feb 11, 2014 by sreemathi.v
edited Mar 24, 2014 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...