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CAT Line and Circle - Some Important results - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Syllabus

Quick Facts

5 Questions around this concept.

Solve by difficulty

EASY

If the line y = mx – 2 is tangent to the circle $x^2+y^2=4$ , then:

m = 1

m = -2

m = 2

m = 0

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Concepts Covered - 1

Line and Circle - Some Important results

In the previous concept we have learnt that three possible cases with a given line and circle.

Now, let us see if we have given the equation of line AB and equation of the circle then what will happen.

Let the equation of line PQ is y = mx + c.

And equation of the circle, with center at origin and radius 'a' is x²+y²=a²

Now put the value of ‘y’ in the equation of circle, we get $\mathrm{x^{2}+(m x+c)^{2}=a^{2} }$

Further simplify it

$\\\Rightarrow\mathrm{ x^2+m^2x^2+2mxc+c^2=a^2}\\\Rightarrow \mathrm{\left(1+m^{2}\right) x^{2}+2 m c x+c^{2}-a^{2}=0}$

Above Equation is a quadratic equation in terms of x² and in the chapter Quadratic equation we have learnt that the nature roots of the quadratic equation depends on the value of D(Discriminant)

Recall

$\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\\\mathrm{D\;(Discriminant)=b^2-4ac}$

So for the quadratic equation $\mathrm{\left(1+m^{2}\right) x^{2}+2 m c x+c^{2}-a^{2}=0}$

$\begin{aligned} \text{D}&=\left(2mc\right)^2-4\left(1+m^2\right)\left(c^2-a^2\right) \\&=a^2+m^2a^2-c^2 \end{aligned}$

So here three cases arise.

Case 1 When D > 0

It means that the equation has two distinct roots, and the line AB intersects the circle at two distinct points,

$\begin{aligned} \text{D}&>0 \\a^2+m^2a^2-c^2&>0 \\a^2\left (1+m^2 \right )-c^2&>0 \\a^{2}&>\frac{c^{2}}{1+m^{2}}\\ a&>\frac{|c|}{\sqrt{1+m^{2}}}=\text{length of perpendicular from (0,0) to the line L} \end{aligned}$

Case 2 When D = 0

It means that the equation has two equal roots and the line AB intersects the circle at one point.

$\begin{aligned} \text{D}&=0 \\a^2+m^2a^2-c^2&=0 \\a^2\left (1+m^2 \right )-c^2&=0 \\a^{2}&=\frac{c^{2}}{1+m^{2}}\\ a&=\frac{|c|}{\sqrt{1+m^{2}}}=\text{length of perpendicular from (0,0) to the line L} \\&=\text{radius of circle} \end{aligned}$

Case 3 When D < 0

It means equation has no real roots, and the line PQ doesn’t intersect the circle at any point,

$\begin{aligned} \text{D}&<0 \\a^2+m^2a^2-c^2&<0 \\a^2\left (1+m^2 \right )-c^2&<0 \\a^{2}&<\frac{c^{2}}{1+m^{2}}\\ a&<\frac{|c|}{\sqrt{1+m^{2}}}=\text{length of perpendicular from (0,0) to the line L} \end{aligned}$