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Find the number of points where Circle x^2+y^2+4x-4y+7=0 intersect the x-axis ?

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

None of these


Answers (1)

best_answer

Intercepts Made by Circle on the Axis 

For \mathrm{x^2+y^2+2gx+2fy+c=0}

Length of x-intercept :\mathrm{2\sqrt{g^2-c}}

 

Now

\\x^2+y^2+4x-4y+7=0\\ x-intercept=2\sqrt{g^2-c}= 2\sqrt{(2)^2-7}= 2\sqrt{-3}

As this is not a real number, so there is no x-intercept made by this circle with x-axis, and hence there is no point of intersection of this circle with x-axis

 

Alternate Method

\\x^2+y^2+4x-4y+7=0\\ \text{Centre of circle (-2,2) and radius r =} \sqrt{4+4-7}=1 \\ \text{Clearly, it won't intersect x-axis axis, so number of points of intersection with x-axis is zero}

Posted by

Devendra Khairwa

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