Today we are going to discuss the *Short Answer Type Questions* of Circle of S N Dey mathematics Class 11.

## S N Dey Math Solutions related to Circle (Ex -3)

Examine whether the equation represents a circle.

Solution.

Hence, by we notice that the radius of the circle being a imaginary number and so, the given equation does not represent a real circle.

Find the equation of the circle passing through and having centre at

Solution.

We know that the general form of equation of the circle is :

where is the centre of the circle given by

Since the circle passing through , we get by and

Hence, by we get, the equation of the circle

Find the centre and radius of each of the following circles :

Solution.

We know that the equation of the circle having centre at and radius unit is

Now, can be written as :

Hence, from and , we get, the centre of the given circle is and the radius is unit.

Solution.

Hence, from we get, the centre of the given circle is and the radius is unit.

Solution.

Hence, from we get, the centre of the given circle is and the radius is unit.

Solution.

Hence, from we get, the centre of the given circle is and the radius is unit.

4. Under what conditions will be the equation of a circle? In that case what will be the co-ordinates of the centre ?

Solution.

will represent an equation of a circle if

For the given equation can be written as follows :

Comparing the equation with the general / standard form of circle, we can say that the centre of the circle is :

Find the radius of the circle which passes through the origin and the points and

Solution.

The general equation of the circle is given by :

Since the circle which passes through the origin, we get ,

So, the equation can be rewritten as

Since the circle passes through the point , we get by

Again, since the circle passes through the point we get by

Hence, the radius of the circle as represented by is given by :

Find the equation of the circle for which the line segment joining the points and and is a diameter.

Solution.

According to the problem, The mid-point of the line segment joining the points and will be the centre of the circle and the centre of the circle is :

The diameter of the circle is the length of the line segment joining the points and and is given by :

and so, the radius of the circle is :

Hence, the required equation of the circle is :

2nd Part :

Solution.

According to the problem, The mid-point of the line segment joining the points and will be the centre of the circle and the centre of the circle is :

The diameter of the circle is the length of the line segment joining the points and and is given by :

and so, the radius of the circle is :

Hence, the required equation of the circle is :

Find the position of the point with respect to the circle whose equation is

Solution.

Comparing the given equation of the circle with the general equation / standard form of circle i.e., with , we get,

Hence, the radius of the circle is :

Again, the centre of the circle is given by :

Now, the distance of the point from the centre of the circle i.e., is :

Hence, from and we get,

and so, the point lies inside the given circle.

The straight line is a tangent to the circle at P ; find the equation of its normal at the same point.

Solution.

Given that straight line is tangent to the circle .

Now, General equation of circle where center of circle is : and radius of circle is :

Comparing the given equation to general equation we get,

So, center of circle .

Now, the equation of the straight line which is perpendicular to the given straight line is : which passes through the point and so,

Hence, the equation of its normal at the same point is given by :

The length of diameter of the circle is find

Solution.

General equation of circle where center of circle is : and radius of circle is :

Comparing the given equation to general equation we get,

So, center of circle .

The radius of the circle :

So, according to the given problem,