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