Circle: Chord, Tangent And Normal

MathsCirclesFor JEE aspirants

INTERSECTION OF A LINE WITH A CIRCLE

Let S=0 be a circle with centre C and radius r ,L=0 be a line and p be perpendicular distance from the centre to the line L=0.

Case I: If p>r i.e. the distance from the centre to the line is greater than the radius then the circle doesn't intersect the line.

Case II: If p=r i.e. the distance from the centre to the line is equal to the radius then the circle will intersect the line in one and only one point i.e. circle touches the line. This is also called the condition of tangency

Case III: If p<r i.e. the distance from the centre to the line is less than the radius then the circle intersect the line in two distinct points.

Diagram being restored — will be back shortly

EQUATION OF CHORD WHOSE MID POINT IS GIVEN

The equation of the chord of the circle S $\equiv$ 0, whose mid point (x₁, y₁) is given by T = S₁.

EQUATION OF TANGENTS

Definition of tangent: The tangent at any point of a circle if defined to be a straight line which meet the circle at that point but being produced, doesn't cut it. The point where tangent meets the curve is called the point of contact.

$∙$ Equation of the tangent to x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is given by xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.

$∙$ The condition that the straight line y = mx + c is a tangent to the circle x² + y²= a²is c² = a² (1 + m²) and the point of contact is (–a²m/c, a²/c) i.e. y = mx $\pm$ $a 1 + m^{2}$ is always a tangent to the circle x² + y² = a² whatever be the value of m.

Illustration 1: Find the locus of the point of intersection of the tangents to the circle x² + y² =4 which are at right angle to each other.

Solution: Let the point of intersection be ( h , k) . Equation of any tangent

y = mx + 2 $1 + m^{2}$

This passes through ( h ,k)

K - mh = 2 $1 + m^{2}$

$\Rightarrow$ (h² - 4 ) m² - 2 khm + k² - 4 = 0. Roots of this equation will represent the slope of the tangents drawn from the point (h, k). Let the roots be m₁ and m₂. Now both the tangents are perpendicular to each other

$\Rightarrow$ m₁.m₂ = - 1 $\Rightarrow$ $\frac{k ^{2} - 4}{h ^{2} - 4} = - 1$ $\Rightarrow$ h² + k² = 8Alternat solution:

Diagram being restored — will be back shortly

From the diagram it is clear that ACBP will always be a square Hence CA² + AP² = CP²

\Rightarrow

h² + k² = 4 + 4Hence the locus is x² + y² = 8

EQUATION OF TANGENT FROM ANY POINT OUTSIDE THE CIRCLE

METHOD 1:Let the equation of given circle be(x – a)² + (y – b)² = r², and we have to find equation of tangent from the point P(x₁, y₁) outside the circle.Equation of any line passing through P(x₁, y₁) is given by y – y₁ = m (x – x₁). For different values of m we get different lines. For some particular value of m this equation will also represent a tangent to circle. To find the value of m apply the condition of tangency.METHOD 2:The joint equation of a pair of tangents drawn from the point P(x₁, y₁)to thecircle x² + y² + 2gx + 2fy + c = 0 is T²=SS_1,where S

\equiv

x²+y²+2gx +2fy + c = 0 ,S₁

\equiv

x₁²+y₁²+2gx₁+2fy₁+c and T

\equiv

xx₁+yy₁+g(x+x₁)+f(y +y₁)+c Illustration 2:Find the equation of tangent. Draw to circle x² + y² – 2x – 2y – 2 = 0 from the point (4, 5) Common Mistake:Student generally use the formula that equation of tangent at(x₁, y₁) on the circlex² + y² + 2gx + 2fy + c = 0 is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0. But they forget that in this case (x₁, y₁) lying on the circle.Solution:Clearly (4, 5) is lying outside the given circle. Hence from the point (4, 5) we can draw two tangents. Equation of any line passing through (5, 4) isy – 4 = m (x – 5).Now to find m we will apply the condition of tangency.

\frac{m - 1 + 4 - 5 m}{m ^{2} + 1} = 2

\Rightarrow

12m² – 24m + 5 = 0

\Rightarrow

\frac{6 \pm 21}{6}

Put these values in (1) to find the required tangency.

EQUATION OF NORMAL

The normal at any point of a curve is the straight line which is perpendicular to the tangent at the point of tangency (point of contact).

Normal of any circle always passes through the centre of the circle.

The equation of the normal to the circle x²+y²+2gx+2fy+c=0 at any point (x₁, y₁) lying on the circle is $\frac{x - x _{1}}{x _{1} + g} = \frac{y - y _{1}}{y _{1} + f}$

LENGTH OF THE TANGENT

The length of the tangent drawn from a point (x₁, y₁) outside the circle S $\equiv$ 0, to the circle is $S_{1}$ .

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