Tangents and Normals MCQ Quiz - Objective Question with Answer for Tangents and Normals - Download Free PDF
Last updated on May 21, 2025
Latest Tangents and Normals MCQ Objective Questions
Tangents and Normals Question 1:
The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to _____.
Answer (Detailed Solution Below) 1 - 768
Tangents and Normals Question 1 Detailed Solution
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.
Tangents and Normals Question 2:
Let circle C be the image of x2 + y2 – 2x + 4y – 4 = 0 in the line 2x – 3y + 5 = 0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α, β), with β < 4, lies on C such that the length of the arc AB is (1/6)th of the perimeter of C, then β - √3α is equal to
Answer (Detailed Solution Below)
Tangents and Normals Question 2 Detailed Solution
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
Tangents and Normals Question 3:
A curve is defined parametrically by the equations \(x=t^2\) and \(y=t^3\). A variable pair of perpendicular lines through the origin \(O\) meet the curve at \(P\) and \(Q\). If the locus of the point of intersection of the tangents at \(P\) and \(Q\) is \(ay^2=bx-1\), then the value of \(a+b\) is
Answer (Detailed Solution Below) 7
Tangents and Normals Question 3 Detailed Solution
According to the question, we have drawn a diagram.
\(\frac{dx}{dt} = 2t \)
\(\frac{dy}{dt} = 3t^2 \)
\(\frac{dy}{dx} = \frac{3t}{2} \)
\(\textbf{The equation of the tangent:} \)
\(y - t^3 = \frac{3t}{2} (x - t^2) \)
\(y - t^3 = \frac{3t}{2} x - \frac{3t^3}{2} \)
\(\textbf{Equation should satisfy the point } (h,k): \)
\(k - t^3 = \frac{3t}{2} h - \frac{3t^3}{2} \)
\(\Rightarrow \frac{t^3}{2} - \frac{3th}{2} + k = 0 \)
\(\Rightarrow t^3 - 3th + 2k = 0 \)
If slope of PQ an QO is m1 and m2 then,
\(m_1 m_2 = -1 \)
\(m_1 = \frac{t_1^3 - 0}{t_1^2 - 0} = t_1, \quad m_2 = \frac{t_2^3 - 0}{t_2^2 - 0} = t_2 \)
Therefore t_1 t_2 = -1
\(\Rightarrow t_3 = 2k \quad \text{[Since, } t_1 t_2 = -1 \text{]} \)
\(\)Substituting 2k in equation \(t^3 - 3th + 2k = 0\),
\((2k)^3 - 3(2k)h + 2k = 0 \)
\(\Rightarrow 8k^3 - 6kh + 2k = 0 \)
\(\Rightarrow k(8k^2 - 6h + 2) = 0 \)
\(\Rightarrow 8k^2 - 6h + 2 = 0 \)
\(\Rightarrow 8y^2 = 6x - 2 \)\(\Rightarrow 4y^2 = 3x - 1 \)
we get,
\(a = 4, \quad b = 3 \)
\(\Rightarrow a + b = 4 + 3 = 7 \)
Tangents and Normals Question 4:
If the circles \(S = x^2 + y^2 - 14x + 6y + 33 = 0 \) and \(S' = x^2 + y^2 - a^2 = 0 \) \(( a \in \mathbb{N} )\) have 4 common tangents, then the possible number of values of a is
Answer (Detailed Solution Below)
Tangents and Normals Question 4 Detailed Solution
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
Tangents and Normals Question 5:
If a tangent to the hyperbola \(\rm x^2-\frac{y^2}{3}=1\) is also a tangent to the parabola y2 = 8x. then equation of such tangent with the positive slope is
Answer (Detailed Solution Below)
Tangents and Normals Question 5 Detailed Solution
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
Top Tangents and Normals MCQ Objective Questions
Find the slope of the tangent of the curve y2 - 3x3 + 2 = 0 at (1, -1)
Answer (Detailed Solution Below)
Tangents and Normals Question 6 Detailed Solution
Download Solution PDFConcept:
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
The number of tangents that can be drawn from (2, 6) to x2 + y2 = 40 is
Answer (Detailed Solution Below)
Tangents and Normals Question 7 Detailed Solution
Download Solution PDFConcept:
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.
The equation of the tangent to the curve x3 + y2 + 3y + x = 0 and passing through the point (2, -1).
Answer (Detailed Solution Below)
Tangents and Normals Question 8 Detailed Solution
Download Solution PDFConcept:
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
If two circles are touching each other externally and with radii 4 and 6 cm respectively. Find the length of common tangent.
Answer (Detailed Solution Below)
Tangents and Normals Question 9 Detailed Solution
Download Solution PDFGiven:
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.
If the line y = mx + c is a tangent to the circle x2 + y2 = a2 then find condition of tangency ?
Answer (Detailed Solution Below)
Tangents and Normals Question 10 Detailed Solution
Download Solution PDFConcept:
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.
For two circles x2 + y2 = 16 and x2 + y2 - 2y = 0, there is / are
Answer (Detailed Solution Below)
Tangents and Normals Question 11 Detailed Solution
Download Solution PDFCONCEPT :
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.
What is the length of the tangent to the circle \(\rm 2x^2+2y^2=8\) from the point (5, 2).
Answer (Detailed Solution Below)
Tangents and Normals Question 12 Detailed Solution
Download Solution PDFConcept:
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.
The centre of the circle passing through the origin and making positive intercepts 4 and 6 on the coordinate axes, lies on the line
Answer (Detailed Solution Below)
Tangents and Normals Question 13 Detailed Solution
Download Solution PDFConcept:
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).
Find the equation of a curve passing through the point ( -2, 3) , given that the slope of the tangent of the curve at any point (x, y) is \(\rm \frac{3x}{y^{2}}\) .
Answer (Detailed Solution Below)
Tangents and Normals Question 14 Detailed Solution
Download Solution PDFConcept:
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.
The length of the curve y = x3/2 over the interval [0,1] will be:
Answer (Detailed Solution Below)
Tangents and Normals Question 15 Detailed Solution
Download Solution PDFConcept:
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) = x3/2
(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
\(\frac{{\partial u}}{{\partial x}} = 9 = \frac{1}{9}\smallint du = dx\)
\(L = \left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\mathop \smallint \limits_4^{13} \left( {\sqrt u } \right).du\)
\(L = \;\left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\left( {\frac{2}{3}} \right){\left[ {{4^{\frac{3}{2}}}} \right]_4}^{13}\)
\(L = \frac{1}{{27}}[\left( {13{)^{\frac{3}{2}}} - {{\left( 4 \right)}^{\frac{3}{2}}}} \right]\)
\(L = \frac{1}{{27}}\left( {{{\left( {13} \right)}^{\frac{3}{2}}} - 8} \right)\)