PROBLEMS ON CIRCLES IN CONIC SECTIONS

Problem 1 :

The equation of the circle passing through (1, 5) and (4, 1) and touching y-axis is

x2 + y2 - 5x - 6y + 9 +λ(4x+3y-19) = 0

where λ is equal to 

a)  0, -40/9   b)  0    3)  40/9    d)  -40/9

Solution :

x2 + y2 - 5x - 6y + 9 +λ(4x+3y-19) = 0

x2+ y2 - 5x + 4λx - 6y + 3λy+ 9 -19λ = 0

x2+ y2 + (4λ - 5)x + (3λ - 6)y + 9 -19λ = 0

The circle touches y-axis, then x = 0

y2 + (3λ - 6)y + 9 -19λ = 0

Since the circle touches y-axis, it will have equal real roots.

b2 - 4ac = 0

(3λ - 6)2 - 4(1)(9 -19λ) = 0

Option a :

λ = 0 and λ = -40/9

these are the values satisfying the equation. So, option a is correct.

Problem 2 :

The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x - 4y = m at two distinct points if 

a)  15 < m < 65   b)  35 < m < 85

3)  -85 < m < -35    d)  -35 < m < 15

Solution :

x2 + y2 = 4x + 8y + 5

x2 + y2 - 4x - 8y - 5 = 0

Comparing the equation with x2 + y2 + 2gx + 2fy + c = 0

2g = -4, 2f = -8

g = -2, f = -4, c = -5

Center of the circle will be (-g, -f) ==> (2, 4)

radius = √g2+f2-c

√22+42-(-5)

√4+16+5

= 5

Length of perpendicular from the point (2, 4) will be lesser than the radius.

So, option d is correct.

Problem 3 :

The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3).

Solution :

Let (h, k) be the center of the circle and r be the radius.

(x - h)2 + (y - k)2 = r

The circle is passing through the point (1, 0).

problems-on-circle-conic-q1

The vertical distance between (1, 0) to the center is k. Then the distance between (h, k) and (2, 3) will be the same because they are radii.

(1 - h)2 + (0 - k)2 = k2

(1 - h)2 = 0

1 - h = 0

h = 1

The circle passes through the point (2, 3). 

(2 - h)2 + (3 - k)2 = r

Applying the value of h

(2 - 1)2 + (3 - k)2 = k2

1 + 9 - 6k + k2 = k2

-6k = -10

k = 10/6

k = 5/3

Problem 4 :

The radius of the circle 3x2 + by2 + 4bx - 6by + b2 = 0

a)  1       b)  3      c)  √10       d)  √11

Solution :

3x2 + by2 + 4bx - 6by + b2 = 0

If the given equation represents equation of circle then, it will have the following characteristics.

i) it is a second degree equation in x and y.

ii) Coefficient of x2 = coefficient of y

iii)  Coefficient of xy = 0

Using the second property, b = 3.

3x2 + 3y2 + 12x - 18y + 9 = 0

Dividing by 3, we get

x2 + y2 + 4x - 6y + 3 = 0

2g = 4 ==> g = 2

2f = -6 ==> f = -3

and c = 3

Finally g = 2, f = -3 and c = 3

radius = √g2 + f2 - c

√22 + (-3)2 - 3

√4 + 9 - 3

√10

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