Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, - Two-dimensional Coordinate Geometry - BITSAT

The circle x^{2}+y^{2}+4x-7y+12=0 cuts an intercept on y-axis is equal to 

  • Option 1)

    7

  • Option 2)

    +4

  • Option 3)

    +3

  • Option 4)

    +1

 

Answers (1)

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

 

y-axis intercept X=0

y^{2}-7y+12=0

y=\frac{7\pm \sqrt{49-48}}{2}

\frac{7\pm1}{2}=4,3

\left | y_{2}-y_{1} \right |=\left | 4-3 \right |=1

 


Option 1)

7

This option is incorrect

Option 2)

+4

This option is incorrect

Option 3)

+3

This option is incorrect

Option 4)

+1

This option is correct

Posted by

Vakul

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 application correction facility begins tomorrow; editable fields
anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20
anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question