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# How are the points $\left( {{\mathbf{1}}, - {\mathbf{3}}} \right),\left( { - {\mathbf{2}}, - {\mathbf{2}}} \right)$ and $\left( {{\mathbf{0}}, - {\mathbf{3}}} \right)\;$ situated with respect to the circle ${x^2} + {y^2} + {\mathbf{2}}x-y + {\mathbf{3}} = {\mathbf{0}}$.

Last updated date: 20th Sep 2024
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Hint : In order to solve this problem put the given coordinates in the given circle and then observe the value, if it is greater than, less than or equal to zero then the point is outside, inside or on the circle respectively.

Given equation is ${x^2} + {y^2} + 2x-y + 3 = 0$
Let $A = \left( {1, - 3} \right),B = \left( { - 2, - 2} \right)\;$ and $C = \left( {0, - 3} \right)$

We will satisfy these points on the given circle if the value is greater than zero then the point lies outside the circle , if it is equal to zero then the point lies on the circle and if less than zero then the point is inside the circle.

For point $A\left( {1, - 3} \right),\,\,\,\,{x^2} + {y^2} + 2x-y + 3 = 1 + 9 + 2 + 3 + 3 = 18 > 0$
Hence point A lies outside the circle

For point $B = \left( { - 2, - 2} \right),\,\,\,\,{x^2} + {y^2} + 2x-y + 3 = 4 + 4 - 4 + 2 + 3 = 9 > 0$
Hence point B lies outside the circle

For point $C = \left( {0, - 3} \right),\,\,\,\,{x^2} + {y^2} + 2x-y + 3 = 0 + 9 + 0 + 3 + 3 = 15 > 0$
Hence point C lies outside the circle.

Note :- Whenever you have asked the location of points inside, on or out of the circle.
Then you have to satisfy the points in the circle, if the value is greater than zero then the point lies outside the circle , if it is equal to zero then the point lies on the circle and if less than zero then the point is inside the circle.