00:01
In this question we are given three non -colinear points and there is only one and only one circle that passes through them and then we are given that the equation may be written in the circle may be rated in the form x squared plus y square plus a x plus b y plus c equal to zero and the three given points are three non -collinear points are okay so the three points are minus 1 .0 and 3 .3 so we can substitute the values of 2 .1 minus 1 .0 and 3 .3 as x and y in three different cases to obtain a relation between a b and c so that we can then solve the three equations in a b and c to find those values so now by taking the first case by substituting the value of 2 .1 as x comma y therefore 2 .1 this is the first case the equation that we would obtain is equal to 2a plus b plus c equal to minus 5 let this be equation number 1 then by substratioting the value of minus 1 comma 0 as x comma y in the second case we would obtain the equation as a plus 0 times b plus c equal to minus 1 let this be equation number 2.
01:37
Then in the third case, by substituting the value of 3 .3 as x comma y.
01:42
Therefore 3 .3 is the third case.
01:45
So the equation obtained would be 3a plus 3b plus c equal to minus 18.
01:54
Let this be question number 3.
01:57
So now we can use the 3 equations and find the values of a, b and c.
02:03
I personally use the kramer's rule as it is quite easier than the other methods.
02:09
Therefore by kramer's rule, by the kramer's rule, now the determinant d would be equal to 211 -1 -1 -0 -1 and then 3 -3 -1.
02:36
The determinant b would be equal to, okay, so the determinant b is nothing but the, it is the determinant having the constant terms in the three given equations so the determinant b would be equal to 5 minus 1 minus 18 so now therefore now we can find the value of da db and dc da is nothing but replacing the first column of the determinant d by determinant b and keeping the second and third columns as it is therefore da is equal to 5 minus 1 minus 18 and then second and third columns are same therefore one zero three then one one one which is the value of the determinant da is equal to minus 35 similarly we can find that value of db which is nothing but replacing the second column of the determinant d with the determinant b and keeping the first and third columns intact that for db is equal to 213 5 minus 1 minus 18 and then 1 1, 1, okay just a minute and then 1, 1...