Tangents and Normals MCQ Quiz - Objective Question with Answer for Tangents and Normals - Download Free PDF

Concept:

Circle Equation and Radius:

• The radius of a circle touching a line can be found using the perpendicular distance from the center to the line.
• The circle equation is \( (x - a)^2 + (y - b)^2 = r^2 \), where \( (a,b) \) is the center and \( r \) is the radius.

Calculation:

Center of circle is \( (a, 0) \).

Radius \( r \) is distance from center to line \( x + y = 3 \):

\( r = \left| \frac{a - 0 - 3}{\sqrt{2}} \right| = \left| \frac{a - 3}{\sqrt{2}} \right| \)

Equation of circle:

\( (x - a)^2 + y^2 = \left( \frac{a - 3}{\sqrt{2}} \right)^2 \)

Circle passes through point \( (-9,4) \):

\( (-9 - a)^2 + 4^2 = \frac{(a - 3)^2}{2} \)

Expanding and simplifying:

\( (a + 9)^2 + 16 = \frac{(a - 3)^2}{2} \)

\( 2(a^2 + 18a + 81 + 8) = a^2 - 6a + 9 \)

\( 2a^2 + 36a + 194 = a^2 - 6a + 9 \)

\( a^2 + 42a + 185 = 0 \)

Factorizing:

\( (a + 37)(a + 5) = 0 \implies a = -37, -5 \)

Calculating radii:

\( r_1 = \left| \frac{-37 - 3}{\sqrt{2}} \right| = \frac{40}{\sqrt{2}} = 20\sqrt{2} \)

\( r_2 = \left| \frac{-5 - 3}{\sqrt{2}} \right| = \frac{8}{\sqrt{2}} = 4\sqrt{2} \)

Absolute difference between squares of radii:

\( |r_1^2 - r_2^2| = |(20\sqrt{2})^2 - (4\sqrt{2})^2| = |800 - 32| = 768 \)

Hence, the correct answer is 768.

Calculation

x² + y² – 2x + 4y – 4 = 0 

Centre (1, –2), r = 3

Reflection of (1, –2) about 2x – 3y + 5 = 0

\(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2\)

x = –3, y = 4

Equation of circle ‘C’

C : (x+3)² + (y– 4)² = 9

\(\ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r}\)

⇒ \(\mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r}\)

⇒ \(\theta=\frac{\pi}{3}\)

(α + 6)² + (β – 4)² = 27

⇒ \(\frac{(\alpha+3)^{2} \pm(\beta-4)^{2}=9}{(\alpha+6)^{2}-(\alpha+3)^{2}=18}\) 

⇒ 6α = –9 

⇒ \(\alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right)\)

∴ \(\beta-\sqrt{3} \alpha\)

\(\left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2}\)

= 4

Hence option 4 is correct

Calculation

Given:

Circle S: \(x^2 + y^2 - 14x + 6y + 33 = 0\)

Circle S': \(x^2 + y^2 - a^2 = 0\), \(a \in \mathbb{N}\)

4 common tangents.

1) Find centers and radii:

S: \(x^2 + y^2 - 14x + 6y + 33 = 0\)

Center: \(C_1(7, -3)\)

Radius: \(r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49 + 9 - 33} = \sqrt{25} = 5\)

S': \(x^2 + y^2 - a^2 = 0\)

Center: \(C_2(0, 0)\)

Radius: \(r_2 = \sqrt{a^2} = a\)

2) Distance between centers:

\(C_1C_2 = \sqrt{(7-0)^2 + (-3-0)^2} = \sqrt{49 + 9} = \sqrt{58}\)

3) For 4 common tangents, \(C_1C_2 > r_1 + r_2\):

⇒ \(\sqrt{58} > 5 + a\)

⇒ \(a < \sqrt{58} - 5\)

⇒ \(a < 7.61 - 5\)

⇒ \(a < 2.61\)

4) Also, \(r_1 \ne r_2\), so \(a \ne 5\).

5) Since \(a \in \mathbb{N}\), possible values of \(a\) are 1 and 2.

∴ The possible number of values of \(a\) is 2.

Hence option 4 is correct

Concept:

Common Tangent to a Hyperbola and a Parabola:

• The given problem asks for the equation of a common tangent to both the hyperbola and the parabola. To find this, we use the properties of the curves and the condition of tangency.
• The general equation of the hyperbola is x²/a² - y²/b² = 1, and for the parabola, it is y² = 8x. A common tangent means the line will touch both curves at exactly one point.
• The tangent to both curves with a positive slope can be derived by considering the slopes of the tangents to each curve and ensuring that they are equal at the point of tangency.

Calculation:

The equation of the hyperbola is:

x²/a² - y²/b² = 1

The equation of the parabola is:

y² = 8x

We are given that the tangent has a positive slope.

Let the equation of the common tangent be in the form y = mx + c, where m is the slope and c is the y-intercept.

For the parabola y² = 8x, the equation of the tangent with slope m is:

y = mx + c (tangent to the parabola)

For the hyperbola, we find the equation of the tangent similarly, ensuring that it satisfies the condition for tangency for both curves.

After solving the system of equations for the tangency condition, we find that the equation of the common tangent with a positive slope is:

y - 2x - 1 = 0

Hence, the correct answer is:

(2) y - 2x - 1 = 0

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given curve y2 - 3x3 + 2 = 0

Differentiating the equation wrt x we get

⇒ 2y\(\rm dy\over dx\) - 9x2 + 0 = 0

⇒ 2y \(\rm dy\over dx\)  = 9x2

Slope at (1, -1)

⇒ 2(-1) \(\rm dy\over dx\) = 9 (1)

⇒ -2\(\rm dy\over dx\) = 9

⇒ \(\rm dy\over dx\) = -4.5

The slope of the tangent (m) = \(\rm {dy\over dx}\)

∴ 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 = \(\rm \frac {dy}{dx}\) 

Calculation:

Given curve x3 + y2 + 3y + x = 0

Differentiating w.r.t x

3x² + 2y\(\rm dy\over dx\) + 3\(\rm dy\over dx\) + 1 = 0

(2y + 3)\(\rm dy\over dx\) = -3x² - 1

\(\rm dy\over dx\) = \(\rm -{3x^2+1\over2y+3}\)

At (2, -1)

\(\rm dy\over dx\) = \(\rm -{3(2)^2+1\over2(-1)+3}\)

\(\rm dy\over dx\) = \(-{12+1\over 1}\) = -13

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 \(\sqrt {1 + {m^2}} \)

Substituting in (2) we see that the equation of a tangent to (1) is 

\(y = mx + a\sqrt {1 + {m^2}} or\,y = mx - a\sqrt {1 + {m^2}} \)

Whatever the value of m.

Observation: It also follows that y = mx +c is a tangent to x² + y² = a² if either:

\(C = a\sqrt {1 + {m^2}} ~or~\,C = - a\sqrt {1 + {m^2}} \)

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

CONCEPT :   

The distance between the centres of two circle is less than the difference of their radii then there is no common tangent.

CALCULATION: 

Equation of first circle x2 + y2   ​= 16 which can be re-written as: x2 + y2  ​= 4²

So, the centre of the first circle (0,0) and radius = 4

Equation of second circle  x2 + y2 - 2y = 0  which can be re-written as: x2 + (y-1)² = 1² 

So, the centre of the second circle is: (0,1) and radius = 1

So, the distance between the centres \(d = \sqrt {{0^2} + {1^2}} = 1\)

The difference between the radii of the two circles = |4 - 1| = 3

As we can see that, the distance between the centres < difference of their radii 

We also know that if the distance between the centres of two circle is less than the difference of their radii then there is no common tangent.

So there is no common tangent between the two given circles.

Hence, option D is the correct answer.

Concept:

The length of the tangent from an external point (x₁, y₁) to the circle x² + y² = a² is given by, \(\rm \sqrt{x_1^2+y_1^2-a^2}\)

Calculation:

Given: equation of circle \(\rm 2x^2+2y^2=8\)

⇒ \(\rm x^2+y^2=4\)

External point (5, 2)

∴ The length of the tangent = \(\rm \sqrt{(5)^2+(2)^2-4}=\sqrt{25+4-4}=5\) units

Hence, option (3) is correct. 

Concept:

The equation of a circle with centre (a,b) and radius r is 

(x - a)2 + (y - b)2 = r2 

The points through which the circle passes satisfy this equation.

Calculation: 

Let the equation of the circle be 

(x - a)² + (y - b)² = r² __(1)

As the circle passes through the origin,

⇒ (0 - a)2 + (0 - b)2 = r2 

⇒  r2 = a2 + b2 

Putting it in (1),

equation of circle becomes (x - a)2 + (y - b)2 = a2 + b2 __(2)

As the circle is making positive intercepts 4 and 6 on the coordinate axes,

⇒ the circle passes through (4,0) and (0,6)

Put (4,0) in (2)

⇒ (4 - a)2 + (0 - b)2 = a2 + b2

⇒ 16 - 8a + a2 + b2 = a2 + b2

⇒ 16 = 8a or a = 2

And putting (0,6) in (2),

⇒ (0 - a)2 + (6 - b)2 = a2 + b2

⇒ a2 + 36 - 12b + b2 = a2 + b2

⇒ 36 = 12b or b = 3

So center of circle = (2,3)

From options, it passes through the line 3x − 4y + 6 = 0

∴ The correct option is (3).

Concept: 

A slope of the tangent to a curve at any point (x, y) is given by \(\rm\frac{\mathrm{d} y}{\mathrm{d} x}\) . 

Calculation: 

We know that slope of the tangent, 

 \(\rm\frac{\mathrm{d} y}{\mathrm{d} x}\) = \(\rm \frac{3x}{y^{2}}\) 

⇒ y² dy = 3x dx 

On integrating both sides , we get

\(\rm \int y^{2}dy = \int 3x \ dx\)

⇒ \(\rm \frac{y^{3}}{3} = \frac{3}{2}x^{2} +C\)          .... (i) 

Given that curve passes through point ( -2, 3 ) ,Thus, substituting x = -2 and y = 3 in (i), we get 

\(\rm \frac{27}{3} = \frac{12}{2} + C\)

 ⇒ C = 3 

Putting C = 3 in (i), we get 

\(\rm \frac{y^{3}}{3} = \frac{3}{2}x^{2} +3\)  is 

The required equation of curve  .

The correct option is 2. 

Concept:

The length (L) of curve y(x) in interval (a,b) is given as:

\(L = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]dx } \)   -----(1)

Calculation:

Given:

y(x) = x^3/2

(a, b) = (0, 1)

\(\frac{{dy\left( x \right)}}{{dx}} = \frac{3}{2}\;{x^{\frac{1}{2}}}\)

From equation (1)

\(I = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + \;\left( {\frac{9}{4}x} \right)} \right]} dx\)

\( = \frac{1}{2}\mathop \smallint \limits_0^1 \sqrt {\left( {4 + 9x} \right).dx} \)

Let, u = 4 + 9x

\(\frac{{\partial u}}{{\partial x}} = 9 = \frac{1}{9}\smallint du = dx\)

\(L = \left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\mathop \smallint \limits_4^{13} \left( {\sqrt u } \right).du\)

\(L = \;\left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\left( {\frac{2}{3}} \right){\left[ {{4^{\frac{3}{2}}}} \right]_4}^{13}\)

\(L = \frac{1}{{27}}[\left( {13{)^{\frac{3}{2}}} - {{\left( 4 \right)}^{\frac{3}{2}}}} \right]\)

\(L = \frac{1}{{27}}\left( {{{\left( {13} \right)}^{\frac{3}{2}}} - 8} \right)\)