Test: Circles: Tangents and Normals (9 July) - JEE MCQ

Given, ∠ACB = 50°
We know that, in a circle, the angle between a chord and a tangent through one of the endpoints of chord is equal to the angle in the alternate segment.
∴ ∠BAT = ∠ABC
⇒ ∠BAT = 50°

Concept:

1. Tangent is perpendicular to the radius at the point of contact.
2. Angle between two lines: The angle θ between the lines having slope m₁ and m₂ is given by 
3. Sum of interior angles of a quadrilateral = 360° 

Calculation:
Given that, tangent to the circle at points A and B is represented by pair of straight lines PA & PB
x² - 3y² - 2x + 1 = 0 -----(1)

From equation (1)
x² - 3y² - 2x + 1 = 0
⇒ (x - 1)² = 3y²
⇒ x - 1 = ± √3y
Therefore, eqation of tangent PA and PB are
x + √3y - 1 = 0    ----(2)
x - √3y - 1 = 0    ----(3)
We know that, slope of line ax + by + c = 0 is -a/b. Therefore,
The slope of line (2) and (3) are -1/√3 and 1/√3 
Therefore, angle b/w tangents PA & PB

We know that,
∠ O + ∠ A + ∠ P + ∠ B = 360º
⇒ ∠ O = 360º - 180º - 60º
⇒ ∠ O = 120º

Concept:
The length of the tangent from an external point P (x₁, y₁) to the circle represented by the equation: x² + y² = a² is given by: 
Calculation:
Given: The distance between the external point and the centre of the circle is 13 cm and radius of the circle is 5 cm.
Let O be the centre of the circle, P be the external point and Q be the point of contact between the circle and the tangent drawn from the point P.
i.e OQ = 5 cm and OP = 13 cm
As we know that, ΔOPQ is a right angle triangle
So, by pythagoras theorem we have
OP² = OQ² + PQ²
⇒ PQ² = 13² - 5² = 144
⇒ PQ = ± 12
∵ PQ is the length of the tangent
⇒ PQ = 12 cm
Hence, option A is the correct answer.

Concept
The distance of point (h,k) from the line ax + by + c = 0 is 
The equation of a circle with centre (h,k) and radius R is, (x - h)² + (y - k)² = R²
Calculation:

Now the equation of the circle with centre (2,-1) and radius  is

Now let the equation of another tangent through origin be y = mx
Put in (1),

 which has only one solution.

⇒ 4m² - 16m + 16 - 10 - 10m² = 0
⇒ 6m² + 16m - 6 = 0
⇒ 2(3m - 1)(m + 3) = 0
⇒ m = -3, (1/3)
⇒ The equation of tangent other than 3x + y = 0 is x - 3y = 0
∴ The correct answer is option (3).

Concept Used:
For any line ax + by + c = 0,
Slope is -a/b and x-intercept is -c/a
Calculation:
For the line, 3x − y + K = 0
a = 3, b = -1 and c = K
⇒ Slope  = -3/-1 = 3 and x-intercept = -K/3
ATQ: Slope = x-intercept
⇒ 3 = -K/3
⇒ K = - 9
Hence, value of K is -9.

Concept:
The slope of the tangent to a curve y = f(x) is m = dy/dx
The slope of the normal = 
Calculation:
Given curve y² - 3x³ + 2 = 0
Differentiating the equation wrt x we get

Slope at (1, -1)

The slope of the tangent (m) = dy/dx
∴ m = -4.5

Concept:
If the point lies inside the circle no tangent can be drawn.
If the point lies on the circle then only one tangent can be drawn
If the point lies outside the circle then a maximum of two tangent lines can be drawn on the circle
Calculation:
Given point (2, 6) and equation of circle is x² + y² = 40
Now, x² + y² - 40 = 0
Put (2, 6) in this equation, we get
2² + 6² – 40 = 4 + 36 – 40 = 40 – 40 = 0
So, the point (2, 6) lies on the circle.
Hence, only one tangent can be drawn.

Concept:
Steps to find the equation of the tangent to the curve:
Find the first derivative of f(x).
Use the point-slope formula to find the equation for the tangent line.
Point-slope is the general form: y - y₁ = m(x - x₁), Where m = slope of tangent = dy/dx 
Calculation:
Given curve x³ + y² + 3y + x = 0
Differentiating w.r.t x

The equation of the tangent is
(y - (-1)) = -13(x - 2)
y + 1 = -13x + 26
y + 13x - 25 = 0

Given:
Two circles are touching each other externally and with radii of 4 and 6 cm
Formula used:
Length of common tangent = √(sum of radius center to center)² - (c₁ -  c₂)²
Calculation:

c₁ = 6 cm, and c₂ = 4 cm
PQ is the length of tangent 
PQ = √(6 + 4)² - (6 - 4)²
⇒ PQ = √(100 - 4)
⇒ PQ = √96
⇒ PQ = 4√6
∴ The length of the tangent is 4√6. 

Concept:
If the line y = mx + c is a tangent to the circle x² + y² = a² 
then c² = ± a²  (1 + m²)
Tangent in terms of m
To find the equation of tangent to the circle x² + y² + = a² in terms of its gradient
Let the equation of circle be x² + y² + = a²      ----(1)
Let the equation of a tangent to (1) be y = mx + c      ----(2)
Solving (1) and (2) simultaneously we have x² + (mx + c)² = a² 
or (1 + m²) x² + 2mcx + ( c² - a²) = 0      ----(3)
(2) is a tangent to (1) if (3) has equal roots.
i.e 4m²c² - 4 (1 + m²) (c² -a²) = 0
[∵ Discriminant = 0]
or m²c² - c² + a² - m²c² + m²a² =0 or c² = a² (1 + m²) or c = ± a
Substituting in (2) we see that the equation of a tangent to (1) is 

Whatever the value of m.
Observation: It also follows that y = mx +c is a tangent to x² + y² = a² if either:

Which is the condition of tangency so the correct answer will be option 1.