Class 12 The Circle Exercise 3.1 Solutions | Pioneer Mathematics Dreamland Publication Pvt. Ltd.

We will use the slope-point form of a straight line to find the equations of tangent and normal to the given circle $\rm x^2 + y^2 = 5$ at (1,2).

The center of the given circle is (0,0) as it is in the standard form.

Given, (x,y) = (1,2)

TANGENT TO CIRCLE

Let the slope of the tangent to the circle be m. We know,

$\rm m = - \frac{x}{y} = - \frac{1}{2}$

$\rm (x_1, y_1) = (1,2)$

Now, the equation of the tangent in slope-point form is

$\rm (y - y_1) = m (x - x_1)$

$\rm or, (y - 2) = - \frac{1}{2} (x - 1)$

Multiplying both sides of the equation by 2, we get,

$\rm or, 2 \cdot (y - 2) = - 2 \cdot \frac{1}{2} (x - 1)$

$\rm or, 2 (y - 2) = - (x - 1)$

$\rm or, 2y - 4 = x + 1$

$\rm \therefore x - 2y + 5 = 0$

Hence, the required equation for the tangent to the given circle at (1,2) is x - 2y + 5 = 0.

NORMAL TO CIRCLE

Let the slope of the normal to the circle be n.

We know that a normal to a circle is perpendicular to a tangent to the same circle. Hence, the product of their slopes is equal to -1.

Mathematically, $\rm m \cdot n = -1$

$\rm or, - \frac{1}{2} \cdot n = -1$

Multiplying both sides of the equation by 2, we get,

$\rm or, - 2 \cdot \frac{1}{2} \cdot n = - 2 \cdot 1$

$\rm \therefore n = 2$

Additionally, a normal to a circle always passes through the center of a circle. So, $(x_1, y_1) = (0,0)$

Now, the equation of the normal in slope-point form is

$\rm (y - y_1) = n (x - x_1)$

$\rm or, (y - 0) = 2( x- 0)$

$\rm or, y = 2x$

$\rm \therefore 2x - y = 0$

Hence, the required equation of the normal to the given circle is 2x - y = 0.

The equation of the given circle is $\rm x^2 + y^2 = 100 \Rightarrow x^2 + y^2 = 10^2$. It is in the standard form, so its center is at (0,0). Its radius is 10 units.

TANGENT TO CIRCLE

The equation of tangent to a standard circle is given by: $\rm xx_1 + yy_1 = a^2$ where $\rm (x_1, y_1)$ are the points through which the tangent passes. And, a is the radius of the circle.

Here, $\rm (x_1, y_1) = (1,2)$ and $\rm a = 10$

Hence,

$\rm x (1) + y (2) = 10^2$

$\rm or, x + 2y = 100$

$\rm or, x + 2y = 100$

$\rm \therefore x + 2y - 100 = 0$

Hence, the required equation of the tangent to the circle is x + 2y - 100 = 0.

NORMAL TO CIRCLE

The equation of normal to a standard circle is given by: $\rm x_1 y = y_1 x$, where $\rm (x_1, y_1)$ are the points through which the normal passes.

Here, $\rm (x_1, y_1) = (1,2)$

Hence,

$\rm (1) y = (2) x$

$\rm y = 2x$

$\rm \therefore 2x - y = 0$

Hence, the required equation of the normal to the circle is 2x - y = 0.

The given circle is in standard form and its equation is $\rm x^2 + y^2 = 169 \Rightarrow x^2 + y^2 = 13^2$. Hence, the circle's center is (h,k) = (0,0) and its radius is 13 units.

TANGENT TO CIRCLE

The equation of tangent to a standard circle is $\rm xx_1 + yy_1 = a^2$.

Here, $\rm (x_1, y_1) = (12, -5)$ and $\rm a = 13$

Hence,

$\rm x (12) + y (-5) = 13^2$

$\rm or, 12 x - 5y = 169$

$\rm \therefore 12x - 5y - 169 = 0$

Hence, the required equation of tangent to the circle is 12x - 5y - 169 = 0.

NORMAL TO CIRCLE

The equation of normal to a standard circle is $\rm x_1 y = y_1 x$

Here, $\rm (x_1, y_1) = (12, -5)$

Hence,

$\rm (12) y = (-5) x$

$\rm \therefore 5x + 12y = 0$

Hence, the required equation of normal to the circle is 5x + 12y = 0.

According to the question, the straight-line x + y =3 intersects the circle at two points, say A $\rm (x_1, y_1)$ and B $\rm (x_2, y_2)$.

Given, the equation of straight-line is

$\rm x + y = 3 \Rightarrow y = x - 3$ – (1)

We substitute this value of y in the given equation of the circle

$\rm x^2 + y^2 - 2x - 3 = 0$

$\rm or, x^2 + (x - 3)^2 - 2x - 3 = 0$

$\rm or, x^2 + (x^2 - 6x + 9) - 2x - 3 = 0$

$\rm or, x^2 + x^2 - 6x - 2x + 9 - 3 = 0$

$\rm or, 2x^2 - 8x + 6 = 0$

$\rm or, 2 ( x^2 - 4x + 3) = 0$

$\rm or, x^2 - 4x + 3 = 0$

The above equation is a quadratic in x. We solve it by the factorization method.

$\rm or, x^2 - (3 + 1)x + 3 = 0$

$\rm or, x^2 - 3x - x + 3 = 0$

$\rm or, x (x - 3) - 1 (x - 3) = 0$

$\rm or, (x - 1) (x - 3) = 0$

Either

$\rm or, (x - 1) = 0$

$\rm \therefore x = 1$

Let this value of $\rm x$ be the value of $\rm x_1$..

Or

$\rm or, (x - 3) = 0$

$\rm \therefore x = 3$

Let this value of $\rm x$ be the value of $\rm x_2$

Put the value of $\rm x_1$ in equation (1) to find the value of $\rm y_1$

$\rm y_1 = x_1 - 3 = 1 -3$

$\rm \therefore y_1 = -2$

Hence, $\rm A(x_1, y_1) = (1, -2)$

Put the value of $\rm x_2$ in equation (1) to find the value of $\rm y_2$

$\rm y_2 = x_2 - 3 = 3 - 3$

$\rm \therefore y_2 = 0$

Hence, $\rm B(x_2, y_2) = (3, 0)$

Now, the required length of the intercepts made by the given line with the given circle is the distance between A and B which is given by the distance formula

$\rm AB = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\rm = \sqrt{ (3 - 1)^2 + (0 - (-2))^2}$

$\rm = \sqrt{ 2^2 + 2^2}$

$\rm = \sqrt{ 4 + 4}$

$\rm = \sqrt{ 8}$

$\rm = 2 \sqrt{2} units$

Hence, the required length of the intercepts made by the straight line $\rm x + y = 3$ with the circle $\rm x^2 + y^2 - 2x - 3 = 0$ is $\rm 2 \sqrt{ 2}$ units.

Given line is $\rm 3x + 4y - 20 = 0$. The formula for the perpendicular distance from a point P (x,y) on the line is $\rm pd = \pm \left ( \frac{3x + 4y - 20}{ \sqrt{3^2 + 4^2} } \right )$, where pd is the perpendicular distance.

We have a standard circle with the equation $\rm x^2 + y^2 = 16$. Its center is at C (h,k) = C (0,0). Similarly, its radius is $\rm 4$ unit.

If the line is tangent to the circle, then the perpendicular distance drawn from the center of the circle to the given line must be equal to the radius of the circle.

Mathematically,

$\rm pd = 4$

$\rm or, \pm \left ( \frac{3x + 4y - 20}{ \sqrt{3^2 + 4^2}} \right ) = 4$

Here, (x,y) = C(h,k) = (0,0)

$\rm or, \pm \left ( \frac{ 3(0) + 4(0) - 20}{ \sqrt{9 + 16}} \right ) = 4$

$\rm or, \pm \left ( \frac{-20}{ \sqrt{25}} \right ) = 4$

$\rm or, \pm \left ( \frac{- 20}{ 5} \right ) = 4$

$\rm or, \pm (-4) = 4$

$\rm \therefore 4 = 4$ which is true.

Hence, we can conclude that the given line touches the circle.

Now, we need to find the point of contact of the circle and the tangent.

The given tangent, $\rm 3x + 4y - 20 = 0$, has a slope of $\rm m = - \frac{3}{4}$. For a tangent to a circle, we also know that its slope is given by $\rm m = - \frac{x_1}{y_1}$.

Solving the two equations, we get,

$\rm - \frac{3}{4} = - \frac{x_1}{y_1}$

$\rm x_1 = \frac{3}{4} y_1$ – (1)

We know,

$\rm y = mx + c$ represents a straight line.

$\rm (y - y_1) = m (x - x_1)$ is also capable of representing the same straight line if $\rm (x_1, y_1)$ lie on the given straight line.

We can rearrange the given equation of tangent as

$\rm 3x + 4y - 20 = 0$

$\rm or, 4y = -3x + 20$

$\rm or, y = - \frac{3}{4} x + 5$ – (2)

Similarly, we can write the equation of the tangent in slope-intercept form as

$\rm (y - y_1) = m (x- x_1)$

where m is the slope of the tangent and point $\rm (x_1, y_1)$ is the required point of contact.

$\rm (y - y_1) = mx - mx_1$

$\rm y = mx + (y_1 - mx_1)$ – (3)

Equations (2) and (3) represent the same tangent, so we can equate the constant terms

$\rm 5 = y_1 - mx_1$

Substituting the values of $\rm m = - \frac{3}{4}$ and $\rm x_1 = \frac{3}{4} y_1$ from (1), we get,

$\rm 5 = y_1 - \left (  - \frac{3}{4} \right ) \cdot \frac{3}{4} y_1$

$\rm or, 5 = y_1 + \frac{3 \cdot 3}{4 \cdot 4} y_1$

$\rm or, 5 = y_1 + \frac{9}{16} y_1$

$\rm or, 5 = y_1 \left ( 1 + \frac{9}{16} \right )$

$\rm or, 5 = y_1 \cdot \frac{16 + 9}{16}$

$\rm or, 5 = y_1 \cdot \frac{25}{16}$

$\rm or, y_1 = \frac{16}{25} \cdot 5$

$\rm \therefore y_1 = \frac{16}{5}$

Put the value of $\rm y_1 = \frac{16}{5}$ in equation (1), we get,

$\rm x_1 = \frac{3}{4} \cdot \frac{16}{5}$

$\rm or, x_1 = \frac{3 \cdot 4}{5}$

$\rm \therefore x_1 = \frac{12}{5}$

Hence, the required point of contact of the tangent to the given circle is $\rm (x_1, y_1) = \left ( \frac{12}{5}, \frac{16}{5} \right )$.

Let line $l_1$ be tangent to the circle $\rm x^2 + y^2 = 25$. The radius of the circle is 5 units. It is given that the angle of inclination $\rm ( \theta )$ of $l_1$ is $\rm 60^o$.

Let m be the slope of the tangent.

We know,

$\rm m = \tan \theta$

$\rm or, m = \tan 60^o$

$\rm \therefore m = \sqrt{3}$

By the slope-intercept form of a straight line, we get,

$\rm y = mx + c \Rightarrow y = \sqrt{3} x + c$

$\rm or, \sqrt{3} x - y + c = 0$

Let the above equation be the equation of the straight line.

Let the perpendicular distance from the center of the circle to the point of intersection on the tangent be pd.

$\rm pd = \pm \left (  \frac{\sqrt{3} x - y + c}{ ( \sqrt{3} )^2 + 1^2} \right )$

Here, (x, y) = (0,0) [Because it is the center of the standard circle]

$\rm pd = \pm \left ( \frac{ \sqrt{3} (0) - (0) + c}{ 3 + 1} \right )$

$\rm pd = \pm \left ( \frac{c}{\sqrt{4}} \right )$

$\rm \therefore pd = \pm \left ( \frac{c}{2} \right ) $

We know that the perpendicular distance from the center of the circle to the point of intersection on the tangent is equal to the radius of the circle.

$\rm pd = r$

$\rm \pm \left ( \frac{c}{2} \right ) = 5$

$\rm or, c = \pm ( 5 \cdot 2 )$

$\rm \therefore c = \pm 10$

Now, substitute the values of m and c in y = mx + c, we get,

$\rm y = \sqrt{3} x \pm 10$ is the required equation of the tangent to the given circle.

Abscissa denotes the x-coordinate of a Point P(x,y). We know that the point of contact of the tangent and the circle will have the coordinates P(1,y). We need to find the corresponding values of y.

It is evident that the point P (1, y) lies on the circle $\rm x^2 + y^2 = 10$. So, we will use this equation to find the corresponding values of y.

$\rm y^2 = 10 - x^2$

$\rm or, y^2 = 10 - (1)^2$

$\rm or, y^2 = 9$

$\rm or, y = \pm \sqrt{9}$

$\rm \therefore y = \pm 3$

So, we get two points of contact P (1, 3) and Q (1, -3). The single tangent can't have two points of contact. Hence, we need to find the equations of two different tangents to the circle.

Let the equation of tangent to the circle at P $\rm (x_1, y_1) =  (1, 3)$ be $\rm l_1$.

The slope of tangent $\rm (m_1) = - \frac{x_1}{y_1} = - \frac{1}{3}$

$\rm \therefore m_1 = - \frac{1}{3}$

Using the slope-point form of an equation, we get,

$\rm ( y - y_1) = m_1 (x - x_1)$

$\rm or, ( y - 3) = - \frac{1}{3} ( x -1)$

$\rm or, 3 (y - 3) = - (x - 1)$

$\rm or, 3y - 9 = -x + 1$

$\rm or, x + 3y = 9 + 1$

$\rm \therefore x + 3y = 10$

Let the equation of the tangent to the circle at P $\rm (x_2, y_2) = (1, -3)$ be $\rm l_2$.

The slope of the tangent $\rm (m_2) = - \frac{x_2}{y_2} = - \frac{1}{-3}$

$\rm \therefore m_2 = \frac{1}{3}$

Using the slope-point form of an equation, we get,

$\rm ( y - y_1) = m_2 (x - x_1)$

$\rm or, ( y - (-3)) = \frac{1}{3} (x - 1)$

$\rm or, 3( y + 3) = (x -1)$

$\rm or, 3y + 9 = x - 1$

$\rm or, x - 3y = 9 + 1$

$\rm \therefore x - 3y = 10$

Hence, the required equations of the tangents to the circle whose abscissa are 1 are found.

Given line is $\rm 2x - y + k = 0$. The formula for the perpendicular distance from a point P (x,y) on the line is $\rm pd = \pm \left ( \frac{2x - y + k}{ \sqrt{2^2 + (-1)^2} } \right )$, where pd is the perpendicular distance.

We have a standard circle with the equation $\rm x^2 + y^2 = 5$. Its center is at C (h,k) = C (0,0). Similarly, its radius is $\rm \sqrt{5}$ unit.

If the line is tangent to the circle, then the perpendicular distance drawn from the center of the circle to the given line must be equal to the radius of the circle.

Mathematically,

$\rm pd = \sqrt{5}$

$\rm \pm \left ( \frac{2x - y + k}{\sqrt{2^2} + (-1)^2} \right ) = \sqrt{5}$

$\rm or, \pm \left ( \frac{2x - y + k}{  \sqrt{5} } \right ) = \sqrt{5}$

Here, (x,y) = C(h,k) = (0,0)

$\rm or, \pm \left ( \frac{ 2(0) - (0) + k }{ \sqrt{5}} \right ) = \sqrt{5}$

$\rm or, \pm \left ( \frac{ k}{ \sqrt{5}} \right ) = \sqrt{5}$

$\rm or, \pm (k) = \sqrt{5} \cdot \sqrt{5}$

$\rm or, \pm (k) = 5$

$\rm \therefore k = \pm 5$

Hence, the required values of k for which the given line may touch the given circle is k = {5, -5}.

A tangent to a circle is a straight line that touches the circle at only one point.

Two tangents, or two straight lines, are perpendicular if the product of their slopes is -1.

We know, the slope of a tangent to a standard circle at the point of contact $\rm (x_1, y_1)$ is given by $\rm m = - \frac{x_1}{y_1}$, where m is the slope of the tangent.

Let $\rm m_1$ and $\rm m_2$ be the slopes of the tangent at (3,4) and (4,-3) respectively.

$\rm m_1 = - \frac{3}{4}$

$\rm m_2 = - \frac{4}{-3} = \frac{4}{3}$

To check if these slopes are perpendicular, we multiply their values.

$\rm m_1 \cdot m_2 = - \frac{3}{4} \cdot \frac{4}{3} = -1$, which is true. They satisfy the condition for perpendicular lines.

Hence, we can conclude that the tangents to the circle at the given points are perpendicular to each other.

Given line is $\rm lx + my + n = 0$. The formula for the perpendicular distance from a point P (x,y) on the line is $\rm pd = \pm \left ( \frac{lx + my + n}{ \sqrt{l^2 + m^2} } \right )$, where pd is the perpendicular distance.

We have a standard circle with the equation $\rm x^2 + y^2 = a^2$. Its center is at C (h,k) = C (0,0). Similarly, its radius is a unit.

If the line is a tangent to the circle, then the perpendicular distance drawn from the center of the circle to the given line must be equal to the radius of the circle.

Mathematically,

$\rm pd = a$

$\rm or, \pm \frac{lx + my + n}{ \sqrt{l^2 + m^2} } = a$

Here, (x,y) = C(h,k) = (0,0)

$\rm or, \pm \frac{ l (0) + m (0) + n}{ \sqrt{l^2 + m^2}} = a$

$\rm or, \pm \frac{n}{\sqrt{l^2 + m^2}} = a$

Squaring both sides to remove the square root, we get,

$\rm or, \left ( \frac{n}{ \sqrt{l^2 + m^2}} \right )^2 = a^2$

$\rm or, \frac{n^2}{ l^2 + m^2} = a^2$

$\rm or, n^2 = a^2 ( l^2 + m^2)$

$\rm \therefore a^2 (l^2 + m^2) = n^2$

Hence, the required condition that the given line is a tangent to the given circle is $\rm a^2 (l^2 + m^2) = n^2$.

The center of the circle is a passing point of the normal to the circle.

Here, the equation of the circle is $\rm x^2 + y^2 + 2gx + 2fy + c = 0$. Its center is $\rm (h,k) = (-g, -f)$.

We know that the equation of the given line is $\rm lx + my + n = 0$. If this line is normal to the given circle, then it must satisfy the values of $\rm (x,y) = (h,k) = (-g,-f)$. So, we replace the values of x and y in the equation of the given line, and we get,

$\rm l (-g) + m (-f) + n = 0$

$\rm or, - gl - fm + n = 0$

Rearranging the given equation, we get,

$\rm \therefore gl + fm = n$

Hence, the required condition that the line $\rm lx + my + n = 0$ is normal to the given circle is $\rm gl + fm = n$.

A line that touches the circle at only one point is a tangent to that circle.

We are given the two-intercept form of a straight line equation $\rm \frac{x}{a} + \frac{y}{b} = 1$. We convert it to a general equation of a straight line by taking the LCM and simplifying it.

$\rm \frac{x}{a} + \frac{y}{b} = 1$

$\rm or, \frac{x}{a} \cdot \frac{b}{b} + \frac{y}{b} \cdot \frac{a}{a} = 1 \cdot \frac{ab}{ab}$

$\rm or, \frac{bx}{ab} + \frac{ay}{ab} = \frac{ab}{ab}$

$\rm or, bx + ay = ab$

$\rm \therefore bx + ay - ab = 0$ is the required general equation of the given line.

If the above line is tangent to the circle $\rm (x - a)^2 + (y - b)^2 = r^2$ with center $\rm (h,k) = (a,b)$ and radius $\rm r = 2$ units, then the radius of the circle is equal to the perpendicular distance between the center of the circle to the straight line.

The formula for the perpendicular distance of a point to the given line is

$\rm d = \left | \frac{bx + ay - ab}{\sqrt{a^2 + b^2}} \right |$

Here, $\rm (x,y) = (h,k) = (a,b)$

$\rm d = \left | \frac{b (a) + a(b) - ab}{ \sqrt{a^2 + b^2} } \right |$

The perpendicular distance ‘d’ must be equal to the radius of the circle ‘r’, so, we get,

$\rm \left | \frac{ab}{\sqrt{a^2 + b^2}} \right | = r$

Squaring both sides

$\rm \left ( \frac{ab}{\sqrt{a^2 + b^2}} \right ) ^2 = r^2$

$\rm or, \left ( \frac{a^2b^2}{a^2 + b^2} \right ) = r^2$

$\rm or, a^2 b^2 = r^2 (a^2 + b^2)$

Dividing both sides of the equation by $\rm a^2 b^2 r^2$, we get,

$\rm or, \frac{a^2 b^2}{a^2 b^2 r^2} = \frac{a^2 r^2}{a^2 b^2 r^2} + \frac{b^2 r^2}{a^2 b^2 r^2}$

$\rm or, \frac{1}{r^2} = \frac{1}{b^2} + \frac{1}{a^2}$

$\rm \therefore \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{r^2}$ is the required condition for the straight line $\rm \frac{x}{a} + \frac{y}{b} = 1$ so that it may touch the circle $\rm (x - a)^2 + (y - b)^2 = r^2$.