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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions- Chapter 24 The Circle – Exercise 24.1 Set 2

### Question 11. A circle of radius 4 units touch the coordinate axis in the first quadrant. Find the equations of its images with respect to the line mirrors x=0 and y=0

Solution:

Centre of the circle is (4, 4)
Equation of the circle is (x-4)2 + (y-4)2 =16
Images of this circle with respect to the line mirrors x=0 and y=0. They have centres at (-4, 4) and (4, -4) respectively.
Required equations are (x+4)2 + (y-4)2 =16 and (x-4)2 + (y+4)2 =16
⇒x2 + y2 +8x -8y +16=0 and x2+y2-8x+8y+16=0

### Question 12. Find the equation of the circles touching y-axis at (0, 3) and making an intercept of 8 units on the x-axis.

Solution:

Centre of the considered circle is (1, 1)
Radius of the considered circle is 1
The circle is rolled along the positive direction of the x-axis. On completing one roll, its centre moves horizontally through a distance equal to its circumference, i.e 2π.
Therefore, the coordinates of the centre of the new circle will be (1+2π, 1).
The required equation of the circle is (x-1-2π)2 +(y-1)2=1

### Question 13. Find the equation of the circles passing through two points on Y-axis a distances of 3 from the origin and having radius 5.

Solution:

The circle passes through the points(0, 3) and (0, -3).
So, (0-h)2 +(3-k)2 = a2 …(1) and (0-h)2 +(-3 -k)2=a2 …(2)
Solving (1) and (2), k=0
Therefore, a2 = 25
From equation (2) we get, h2 +9 =25 ⇒ h=±4
So, the required equation is (x±4)2 +y2 = 25 ⇒ x2 ± 8x +y2 -9 = 0

### Question 14. If the lines 2x -3y = 5 and 3x – 4y = 7 are the diametersof a circle of area 154 square units, then the equation of the circle.

Solution:

πr2 = 154 ⇒ r2 = 49
The point of intersection of both the lines will be the centre of the circle.
Solving 2x -3y = 5 and 3x – 4y = 7 we get, x=1 and y=-1
Putting the values in the standard equation, we get (x-1)2 +(y+1)2 =49
⇒ x2 + y2 – 2x + 2y -47 = 0

### Question 15. If the line y = √3 x + k touches the circle x2 +y2 = 16, then find the value of k.

Solution:

From the given equation of circle we can conclude the centre is at (0, 0) and radius is 4.
The perpendicular distance from the centre of the circle to the tangent y = √3 x + k is equal to the radius of the circle.
Using the formula of perpendicular distance of point from a line, we get k=±8.

### Question 16. Find the equation of the circle having (1, -2) as its centre and passing through the intersection of the lines 3x +y = 14 and 2x + 5y = 18

Solution:

Solving 3x+y=14 and 2x+5y=18 we get
x=4 and y=2
The radius is equal to the distance between (1,-2) and (4,2)
r=√((4-1)2+(2+2)2)
=√(9+16)
=5
Putting values in the standard equation,
⇒(x-1)2+(y+2)2=25
⇒x2+y2-2x+4y-20=0

### Question 17. If the lines 3x-4y+4=0 and 6x-8y-7=0 are tangents to a circle, then find the radius of the circle.

Solution:

Given, 3x-4y+4=0 and 6x-8y-7=0
⇒y=3/4x+1 and y=3/4x-7/8
The slope of both the lines is equal so, both the lines are parallel.
Using the formula of the distance between the parallel lines we get, 3/2
The radius is equal to the half of the distance between the parallel lines (diameter of the circle).
Therefore , the radius is given by 3/4.

### Question 18. Show that the point (x, y) given by x=2at/ 1+t2 and y=a(1-t2/1+t2) lies on a circle for all real values of t such that -1<=t<=1 where a is any given real number.

Solution:

Square and add x=2at/1+t2 and y=a(1-t2/1+t2), we get
x2+y2=(2at/1+t2)2+a2(1-t2/1+t2)2
⇒x2+y2=(4a2t2+a2-2a2t2+a2t2)/(1+t2)2
⇒x2+y2=a2((1+t2)2/((1+t2)2))
⇒x2+y2=a2
The above equation represents the equation of a circle, hence points (x, y) lie on the circle.

### Question 19. The circle x2+y2-2x-2y+1=0 is rolled along the positive direction of x-axis and makes one complete roll . Find its equation in new position.

Solution:

Centre of the circle is (1,1)
Radius of the circle is 1
The circle is rolled along the positive direction of the x-axis. On making one complete roll, its centre moves horizontally a distance equal to its circumference i.e. 2π
So, the coordinate of the centre of the new circle will be (1+2π,1)
Therefore, the required equation of the circle is (x-1-2π)2 +(y-1)2=1

### Question 20. One diameter of the circle circumscribing the rectangle ABCD is 4y=x+7. If the coordinate of A and B are (-3,4) and (5,4) respectively, find the equation of the circle.

Solution:

The centre of the circle lies on the line 4y=x+7 and circle passes through A(-3,4) and B(5,4).
Slope of the segment joining A and B is zero .
So, the slope of the perpendicular bisector of AB is not defined.
The perpendicular bisector of AB will be parallel to the y-axis and pass through ((-3+5/2), (4+4/2)) = (1,4)
Equation of the perpendicular bisector is x =1
Intersection point of the perpendicular bisector and 4y=x+7 is (1,2)
The required equation of the circle is x2+y2-2x-4y-15=0

### Question 21. If the line 2x-y+1=0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x+y-9=0. Find the equation of the circle.

Solution:

Let coordinates of the centre be (t, 9-t) and the radius be a.
Using the formula of perpendicular distance of line from a point, we get
a = |(3t-8)/√5| ⇒ a2 =((3t-8)/√5)2 …(1)
From the deductions, the equation of the circle is (x-t)2+(y-(9-t))2=a2 …(2)
The circle passes through (2, 5). Substituting values we get, t=6
Substituting t in (1), a2 =(10/√5)2
Putting the values found in (2), the required equation is,
(x-6)2+(y-3)2=20

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