**Chapter 6 Circle Miscellaneous Exercise 6**

## Chapter 6 Circle Miscellaneous Exercise 6

**(I) Choose the correct alternative.**

**Question 1.Equation of a circle which passes through (3, 6) and touches the axes is**

**Question 2.If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.**

**Question 3.Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3** = 0.

**Question 4.The equation(s) of the tangent(s) to the circle x ^{2} + y^{2} = 4 which are parallel to x + 2y + 3 = 0 are(A) x – 2y = 2(B) x + 2y = ±2√3(C) x + 2y = ±2√5(D) x – 2y = ±2√5Answer:**

(C) x + 2y = ±2√5

**Question 5.If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.**

Hint:

Tangents are parallel to each other.

The perpendicular distance between tangents = diameter

**Question 6.The area of the circle having centre at (1, 2) and passing through (4, 6) is**
(A) 5π

(B) 10π

(C) 25π

(D) 100π

**Answer:**

(C) 25π

Hint:

**Question 7.If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.**

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**Question 9.A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is**

Hint:

**Question 10.**

**(II) Answer the following:**

**Question 1.**
and radius of the circle

**Question 2.**

**Question 3.Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.Solution:**

Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0.

x + 3y = 0

⇒ x = -3y ……..(i)

2x – 7y = 0 ……(ii)

Substituting x = -3y in (ii), we get

⇒ 2(-3y) – 7y = 0

⇒ -6y – 7y = 0

⇒ -13y = 0

⇒ y = 0

Substituting y = 0 in (i), we get

x = -3(0) = 0

Point of intersection is O(0, 0).

This point O(0, 0) lies on the circle.

Let C(h, k) be the centre of the required circle.

Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle.

∴ x = h, y = k

∴ Equations of lines become

h + k = -1 ……(iii)

h – 2k = -4 …..(iv)

By (iii) – (iv), we get

3k = 3

⇒ k = 1

Substituting k = 1 in (iii), we get

h + 1 = -1

⇒ h = -2

∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0).

**Question 4.Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.Solution:**

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B.

∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).

Since ∠BOA is a right angle.

AB represents the diameter of the circle.

**Question 5.Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.Solution:**

Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be

+ 2gx + 2fy + c = 0 …….(i)

For point (9, 1),

Substituting x = 9 andy = 1 in (i), we get

81 + 1 + 18g + 2f + c = 0

⇒ 18g + 2f + c = -82 …..(ii)

For point (7, 9),

Substituting x = 7 andy = 9 in (i), we get

49 + 81 + 14g + 18f + c = 0

⇒ 14g + 18f + c = -130 ……(iii)

For point (-2, 12),

Substituting x = -2 and y = 12 in (i), we get

4 + 144 – 4g + 24f + c = 0

⇒ -4g + 24f + c = -148 …..(iv)

By (ii) – (iii), we get

4g – 16f = 48

⇒ g – 4f = 12 …..(v)

By (iii) – (iv), we get

18g – 6f = 18

⇒ 3g – f = 3 ……(vi)

By 3 × (v) – (vi), we get

-11f = 33

⇒ f = -3

Substituting f = -3 in (vi), we get

3g – (-3) = 3

⇒ 3g + 3 = 3

⇒ g = 0

Substituting g = 0 and f = -3 in (ii), we get

18(0) + 2(-3) + c = – 82

⇒ -6 + c = -82

⇒ c = -76

= 36 + 100 – 60 – 76

= 0

= R.H.S.

∴ Point (6,10) satisfies equation (vii).

∴ the given points are concyciic.

**Question 6.**
When x = -5,

**6 = -4**

y = -10 +

y = -10 +

∴ Point of intersection in B (-5, -4).

**Question 7.**

the equation of the tangent at (-1, 1) is

⇒ x(-1) + y(1) – 2(x – 1) – 1(y + 1) – 4 = 0

⇒ -3x – 3 = 0

⇒ -x – 1 = 0

⇒ x = -1

∴ x = -1 is the tangent to the given circle at (-1, 1).

**Question 8.**

**Question 9.Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ.Solution:**

The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.

**Question 10.**

**Question 11.Find the lengths of the intercepts made on the co-ordinate axes, by the circles.**

substitute y = 0 and get a quadratic equation in x, whose roots are, say, x

_{1}and x

_{2}.

These values represent the abscissae of ends A and B of the x-intercept.

Alternate Method:

(ii) Given equation of the circle is

∴ Length ofy-intercept = 15 units

**Question 12.Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent.**

Equation of common tangent is

∴ P divides C_{1} C_{2} internally in the ratio r_{1} : r_{2} i.e. 1 : 2

∴ By internal division,

Point of contact = (1, 2)

Equation of common tangent is

( – 4x – 10y + 19) – ( + 2x + 8y – 23) = 0

⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0

⇒ -6x – 18y + 42 = 0

⇒ x + 3y – 7 = 0

**Question 13.Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent.**

**Question 15.**

**Question 16.**

**Question 18.**

**Question 19.**

**Question 20.**

Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

**Question 21.**

**Question 22.**

**Question 23.**

**Question 24.**

**Question 25.**

Alternate method:

**Question 26.**