# Circle | Part-8 | S N Dey

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

##### Circle Related Long Answer Type Questions (11-20) | S N Dey Mathematics

11. Show that the circle touches the co-ordinate axes. Also find the equation of the circle which passes through the common points of intersection of the above circle and the straight line and which also passes through the origin.

Solution.

The equation of the given circle can be written as

From we get the centre of the circle is and radius unit.

Now, if , we get from the given equation of the circle,

As we have just one point common between axis and circle, it means it touches
axis at

Similarly, if , we get from the given equation of the circle,

As we have just one point common between axis and circle, it means it touches
axis at

Hence, the given circle touches the two axes.

Now, the equation of the circle passing through the intersection of the given circle and the straight line is

Since the circle passes through the point , we get from

Hence, the equation of the circle is

12. A circle through the common points of the circles and has its centre on the line Find the centre and radius of the circle.

Solution.

Any circle through the common points of the circles and can be written as

Comparing with , we get

So, centre of the circle is

Since the centre of the circle lies on the line ,

Hence, the centre of the circle

Also, the radius of the circle is

13. The circle and the line intersect at and . Find the equation of the circle on as diameter.

Solution.

The equation of the given straight line

The equation of the given circle is

From we get,

From and we get,

Now, by we get for

Similarly, for

So, the equation of the circle on as diameter is given by

14. Find the equation to the circle described on the common chord of the circles and as a diameter.

Solution.

Two given equations of circles are

So, by we get the equation of common chord which is

Equation of any circle through the intersection of two given circles are

From we get the centre of circle which lies on the common chord

Now, putting the value of in we get

15. Find the equation to the locus of the mid-points of chords drawn through the point on the circle

Solution.

Let be the chord passing through the point

Any point on the circle can be written as

If is the co-ordinates of mid-point of , then

Then from we get,

Hence, by we can say that the equation to the locus of the mid-points of chords is

16. A circle passes through the origin and intersects the co-ordinate axes at and . If the length of the diameter of the circle be unit, then find the locus of the centroid of the triangle

Solution.

Any circle passing through the origin can be written as

Let

Since the circle passes through the points so

If be the centroid of , then

So, by we get,

the locus of the centroid of the triangle is

17. Find the equation of a circle circumscribing the triangle whose sides are and If vary so that , find the locus of the centre of the circle.

Solution.

18.Find the area of the equilateral triangle inscribed in the circle

Solution.

The equation of the given circle is

The centre of the circle is and radius

Now, from the figure , we note that in

So, the area of the equilateral triangle inscribed in the circle is

19. Find the area of the equilateral triangle inscribed in the circle

Solution.

Comparing the given circle with the standard form of the circle we get,

Now, from the figure () we notice that

So, the length of the side of is

So, the area of the equilateral triangle inscribed in the circle

20. Prove analytically that the straight line joining the middle point of a chord of a circle of a circle with the centre is perpendicular to the chord.

Solution.

To prove : Given that

From and , we get

Hence,