In the previous article , we discussed Very Short Answer Type Questions. In this article, we will discuss 10 Short Answer type Questions from Chhaya Mathematics , Class 11 (S N De book ).

##### Short Answer Type Questions of Circle, S N Dey Mathematics, Class 11

1. Find the equation of the circle whose centre is and which passes through the centre of the circle

Solution.

The equation of the circle with centre and radius can be written as :

By the given question, Putting these values of in we get from

Now, comparing the given the circle with the general equation of circle we get ,

Hence, the centre of the given circle is :

Now, since the circle as represented by passes through we get,

Hence, from we get the required equation of circle which is :

2. Find the equation of the circle concentric with the circle and passing through the point

Solution.

Now, comparing the given the circle with the general equation of circle we get ,

So, the centre of the circle , is :

Also, the equation of the circle with the centre is given by

Since the circle represented by passes through the point we get,

Hence, from we get the required the equation of the circle which is

Hence, the equation of the circle concentric with the circle and passing through the point is

3. Find the equation of the circle concentric with the circle and passing through the centre of the circle

Solution.

Comparing the given the circle with the general equation of circle we get ,

Since we need to find the equation of the circle concentric with the circle i.e. with the centre , the equation of the circle with radius can be written as

Now, we have to find the centre of the circle

Comparing the given the circle with the general equation of circle we get ,

So, the centre of the circle :

Now, the circle represented by passes through the point so

Finally, from we get the equation of the circle which is

Hence, the required equation of the circle is

4. Find the equation of the straight line which passes through the centre of the circle and is perpendicular to the straight line

Solution.

The equation of the straight line which is perpendicular to the straight line can be written as : being an arbitrary constant.

Now, we need to find the centre of the circle

Comparing the given the circle with the general equation of circle we get ,

Hence, the centre of the circle is given by

Now, since the straight line passes through the point we have

Finally putting the value of in , we get the required equation of the straight line which is

5. Find the equation of each of the circles passing through the points

Solution.

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

Since the circle in passes through the points we get

Now, putting the value of in we get,

Hence, by putting the values of in we get the required equation of circle which is

6.Show that the points and are concylic ; find the equation of the circle on which the points lie and the co-ordinates of the centre of the circle.

Solution.

The general equation of circle

Since the points and are concylic, we get from

Subtracting from we get

Subtracting from we get

From , we get

From we get,

Putting the value of in we get,

Hence, the equation of the required circle is

Comparing with the general equation of circle we get ,

Hence, the co-ordinates of the centre of the circle is

7. Prove that the centres of the circles and lie on a line; find the equation of the line on which they lie.

Solution.

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

The centre of the circle represented by is

Now, comparing with we get,

So, the centre of the circle represented by is

Similarly, the centre of the circle is and the centre of the circle is

Now, the equation of the line passing through is :

Now, the putting in we get,

Hence, the point satisfies the equation

So, the centres of three circles lie on the line

8. Show that the centres of the following circles lie on a line and their radii are in A.P. :

Solution.

The equation of the first circle can be written as so that we can say the centre of the first circle is and radius is unit.

The equation of the second circle is so that the the centre of the second circle is :

Also, the radius is :

The equation of the third circle is so that the the centre of the third circle is :

Also, the radius is :

Clearly, the radii of three circles are respectively and so and so the radii are in A.P.

Again, the slope of the straight line joining the points and is :

Also, the slope of the straight line joining the points and is :

Hence, from and we can conclude that the centres of the following circles lie on a line.

9. Find the equation of the circle which passes through the origin and cuts off intercepts unit and unit from and -axes respectively. Find the equation of that diameter of the circle which passes through the origin.

Solution.

Suppose that the circle which passes through the origin , cuts off intercepts unit and unit from and -axes at the point and respectively. Then,

The circle passes through the origin and

is the diameter of the circle, the equation of the circle having diameter with can be written as

Now , we need to find the equation of the diameter of this circle which passes through the .

The centre of the circle is

Now, the diameter of this circle passes through

So, the equation of the straight line which passes through is given by :

10. Find the equation of the circle circumscribing the triangle formed by the straight line with the axes of co-ordinates. What is the diameter of the circle ?

Solution.

Let the given straight line intercepts the axes of co-ordinates at the points and respectively.

Now,

From we notice that the straight line cuts off units from the axis and units from axis.

Now, Clearly, is a right angled triangle and is the mid point of which is a diameter of circumscribed circle with extremities

Now, the equation of the circle having diameter with can be written as

Also, diameter