no error circles paper 1

PAPER – 1(circle)

SINGLE CHOICE CORRECT

1. The set of values of ‘c’ so that the equations and have no solution is

(a) (b)

(c) (d)

2. Two points P and Q are taken on the line joining the points A (0,0) and B (3a,0) such that

AP = PQ = QB. Circles are drawn on AP, PQ and QB as diameters. The locus of the points, the sum of the squares of the tangents from which to the three circles is equal to b2, is

(a) (b)

(c) (d)

3. Let be the equation of a circle. If has equal roots and has

roots then the centre of the circle is

(a) (2, 29/10) (b) (29/10, 2) (c) (-2, 29/10) (d) none of these

4. The locus of the point of intersection of the tangents to the circle at points whose parametric angles differ by is

(a) (b)

(c) (d)

5. If and a line through cuts the circle in A and B, then

is equal to

(a) 4 (b) 8 (c) 16 (d) 32

6. The locus of the centers of the circles which cut the circles and

orthogonally is

(a) (b)

Head Office: Andheri: 26245223 MUMBAI / DELHI/ AKOLA / KOLKATA / LUCKNOW

(c) (d)

7. The common chord of and subtends an angle at the origin equal to

(a) (b) (c) (d)

8. If the distances from the origin of the centers of three circles are in

GP, then the lengths of the tangents drawn to them from any point on the circle are in

(a) AP (b) GP (c) HP (d) none of these

9. If and the line touches a fixed circle, then

(a) the centre of the circle is at the point (4, 0)

(b) the radius of the circle is equal to

(c) the circle passes through origin

(d) none of the above

10. If for all and , and , then the equation of the

circle having and as the ends of its one diameter is

(a) (b)

(c) (d)

MULTIPLE CORRECT QUESTIONS

11. The tangents drawn from the origin to the circle are perpendicular, if

(a) (b) (c) (d)

12. If is the angle subtended at by the circle , then


(a) (b)

(c) (d)

13. The equation of a common tangent to the circles and

is

(a) (b) (c) (d)

14. A and B are two points on the circle which are farthest and nearest respectively from the point (7,2), then

(a) (b)

(c) (d)

15. The equations of four circles are . The radius of a circle touching all the four circles is

(a) (b) (c) (d)

16. The equation of a circle is . The locus of the intersection of perpendicular tangents to the

circle is the curve and the locus of the intersection of perpendicular tangents to the curve is the

curve . Then

(a) is a circle (b) the area enclosed by the curve is

(c) and are circles with the same centre (d) none of above

17. The equation of the tangents drawn from the origin to the circle are

(a) (b) (c) (d)


PASSAGE TYPE

18. If and we have to find equation of circle having is a tangent and we can

adjust given condition as or

Centre of circle = (4, 0) and radius = 3

1. If , then equation of the circle having is a tangent is

(a) (b)

(c) (d)

2. If and if S be the equation of the circle having is a tangent when the equation of director circle of S is

(a) (b)

(c) (d)

INTEGER TYPE

1. A region in the x-y plane is bounded by the curve and the line . If the point

lies in the interior of the region, then . The value of is

2. The interval of value of for which the line bisects two chords drawn from a point

to the circle is . Then the value

of 'c' is


ANSWER KEY

1. (d) 2. (b) 3. (b) 4. (d) 5. (c) 6. (b) 7. (d)

8. (b) 9. (b) 10. (a) 11. (a,b,c) 12. (b,d) 13. (b,c)

14. (b,d) 15. (a,c) 16. (a,c) 17. (a,c) 18. 1-(b),2-(c)

Integer

1) 2

2) 2


SOLUTION

1. Since and both are symmetrical about y-axis for

Equation of tangent to circle

Parallel to is

for no solution

2. Since AP = PQ = QB . the coordinates of P are (a, 0)


and of Q are (2a, 0) the centre of the circles on AP, PQ and QB as diameters are respectively

and and the radius of each one of them is .

Hence, the equations of the circles with centre and are respectively.

And

So that, if S(h, k) be any point on the locus, then

3. Let

Have equal roots,

Then

Have roots 4/5,5


4. Circle is

Equation of tangent at is

…(i)

…(ii)

Squaring and adding Eqs. (i) and (ii), then we get

5. Given,


Equating the coefficients of and , we get

Equation of line is then point

lies on the circle

Alternative Method:

6. Let circle according to question

...(i) Head Office: Andheri: 26245223 MUMBAI / DELHI/ AKOLA / KOLKATA / LUCKNOW

...(ii)

Subtracting Eqs. (ii) from (i),

7. Equation of common chord is

8. Centres of circles are

If

are in GP

…(i)

Let any point on

Length of tangents are


[from Eq. (i)]

Hence are in GP.

9. Let the circle be

Comparing the coefficients of similar terms

10. …(i)

In Eq. (i), Put

Now in Eq. (i),


Equation of circle in diametric form is

11.

…(i)

Since tangents are perpendicular , then locus of point of intersection of tangents is director circle.

Director circle of (i) is

…(ii)

Since point of intersection of tangent is (0,0), then from Eq.

(ii)

12.


13. Centres and radii of the given circles are

For common tangent tangent length of perpendicular from centre on tangent = radius

Of centre then (b) and (c) are correct.

14. Slop of

If

Equation of

lie on circle,


15. Radius of outer circle

Radius of outer circle = OR + RQ

16. is the director circle of

Equation of is

Again is the director circle of , Hence the equation of is

17. The given equation is


Tangents are x = 0

and y = x tan

18. 1.

Centre = (4,0) and radius = 4

Equation of circle is

2.


Centre = (3,4) and radius = 5

Equation of circle is

Equation of director circle is


no error circles paper 1

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