 Hello students, let's work out the following problem. It says prove that the plane x plus 2 y minus z is equal to 4. Intersects the sphere, x square plus y square plus z square minus x plus z minus 2 is equal to 0 in a circle of radius unity and find the center of the circle. So let's now move on to the solution. We are given that the plane intersects the sphere and we have to prove that the circle which is obtained by intersecting the sphere by the plane has radius unity and we also have to find the center of the circle. That is, we have to find the coordinates of the point A and we have to find the length of AP and we have to prove that AP is 1 unit. So let's now move on to the solution. The given sphere is x square plus y square plus z square minus x plus z minus 2 is equal to 0. Now, the general equation of the sphere is x square plus y square plus z square plus 2 fx plus 2 gy plus 2 hz plus dz equal to 0 and the center of this sphere is minus f minus g minus h and the radius is given by this formula under the root x square plus g square plus x square minus d. So you need to remember this formula and the general form of the equation of the sphere. Now, from this center of the sphere is 1 by 2 0 since the coefficient of y here is 0 and minus 1 by 2 and the radius is given by under the root 1 by 2 square plus 0 square plus minus 1 by 2 square minus of minus 2 is plus 2. So, this is equal to 1 by 4 plus 1 by 4 plus 2. So, this is equal to taking a LCM we have and simplifying we have 10 upon 4 which is equal to under the root 5 by 2. So, this is the radius of the sphere. Now, we have to find the center of the circle. Now, the plane is x plus 2y minus z minus 4 is equal to 0. Now, the direction ratios of the plane, direction ratios of the normal to the plane are 1 2 minus 1 the coefficient of x, y and z here. Now, the center of the circle is the foot of perpendicular from the center of the sphere to the plane right. So, now the equation of line OA will be x minus 1 by 2 upon 1 is equal to y minus 0 upon y minus 0 upon 2 is equal to z minus minus 1 by 2 that is z plus 1 by 2 upon minus 1. As we know that the equation of the line passing through the point a b c and having direction ratios as l m n is given by x minus a upon l y minus b upon m z minus c upon n. And here the line OA passes through the center O and it has direction ratios as 1 2 minus 1. So, this is the equation of the line OA let it be equal to lambda. So, now we have x is equal to lambda plus 1 by 2 y is equal to 2 lambda z is equal to minus lambda minus 1 by 2 that means any point on the line is of the form lambda plus 1 by 2 2 lambda minus lambda minus 1 by 2. Now, since the point a which is the center of the circle lies on the plane it will satisfy the equation of the plane that means this point will satisfy the equation of the plane as center lambda plus 1 by 2 2 lambda minus lambda minus 1 by 2 lies on the plane it will satisfy the equation of the plane. So, we have equation of plane is x plus 2 y now x is lambda plus 1 by 2 2 y y is 2 lambda minus z z is minus lambda minus 1 by 2 and this can be written as minus of lambda plus 1 by 2. So, this becomes and this is equal to 0. So, we have lambda plus 1 by 2 plus 4 lambda plus lambda plus 1 by 2 is equal to 0. So, we have lambda plus 4 lambda plus lambda is 6 lambda 1 by 2 plus 1 by 2 is 1 and this is equal to 0, but in equation of equation of plane was x plus 2 y minus z minus 4 is equal to 0. So, here we have minus 4. So, this implies 6 lambda minus 3 is equal to 0 and this implies lambda is equal to 1 by 2. So, the coordinates of the point a lambda plus 1 by 2 and lambda is 1 by 2. So, this is 1 by 2 plus 1 by 2 2 lambda and minus lambda minus 1 by 2. So, the center a is 1 1 minus 1. Now, we have to prove that the radius of the circle is 1 unit. So, we will prove that radius of this circle is 1 unit by using the Pythagoras theorem. Now, here the radius of the sphere is root 5 by 2 and now we need to know the distance between o and a that is the perpendicular distance from the center to the plane. So, let p 1 be the perpendicular distance from o to the plane. Now, the perpendicular distance from the point x 1 y 1 z 1 to the plane a x plus b y plus c z plus d is equal to 0 is given by the formula mod of a x 1 plus b y 1 plus c z 1 plus d upon under the root a square plus b square plus c square. So, now, the plane is x plus 2 y minus z minus 4 is equal to 0 and the center is center of the sphere is 1 by 2 0 minus 1 by 2. So, p 1 is given by the formula mod of a x 1 that is 1 into 1 by 2 plus b here b is 2 y 1 which is 0 plus c z 1. Now, here c is minus 1 z 1 is minus 1 by 2 here d is minus 1 by 2. So, this is minus 4 upon a square b square under the root a square plus b square plus c square. Here a is 1, v is 2, c is minus 1 and this is equal to mod of minus 3 by under the root 6 which is equal to 3 by under the root 6. Assume that p 1 is the perpendicular distance from the center o to the plane this. So, p 1 is the distance of o and a. So, this is o a. Now, we have to prove that a p is 1 unit. So, here we will use Pythagoras theorem. a p is under the root of o p square minus o a square. Now, o p is the radius of the sphere which is root 5 by 2 minus o a square which is 3 by root 6 square. So, this is equal to under the root 5 by 2 minus 9 by 6. So, this is again equal to under the root 5 by 2 minus 3 by 2. So, this is equal to under the root 2 by 2 which is equal to root 1 which is equal to 1. Hence, proved 1 unit and the center of the circle is 1 1 minus 1. So, this completes the question and the session. Bye for now. Take care. Have a good day.