Question

A straight line :$x=y+2$ touches the circle $4(x^2 + y^2) =r^2$ . The value of r is??

please explain with proper steps.i am not able to understand​

Answer 1

Solution

A straight line $x-y-2=0$ and $x^2+y^2=\dfrac{r^2}{4}$ touches each other if the distance from the center is equal.

The circle is centered at the origin. Let's find the distance between the origin and the line.

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Distance

• $O(0,0)$
• $x-y-2=0$

The distance between a point and a straight line is $d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2} }$.

$d=\dfrac{|0-0-2|}{\sqrt{1^2+(-1)^2} }$

$=\dfrac{2}{\sqrt{2} }=\boxed{\sqrt{2}}$

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Line and a Circle

Relation between a straight line and a circle

• Does not touch if $d>r$
• Touches if $d=r$
• Meets twice if $d<r$

Where,

• $d$ is the distance from a straight line to the center
• $r$ is the radius

So we need $d=r$. This is equivalent to $\boxed{d^2=r^2}$ since $d,r>0$.

• The radius squared is $\dfrac{r^2}{4}$.
• The distance squared is $2$.

$\implies \dfrac{r^2}{4} =2$

$\implies r^2=8$

$\implies r=2\sqrt{2} \ \mathrm{(since\ r>0)}$

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When the value of $r$ is $2\sqrt{2}$, the line and the circle touch each other. This is the required answer.

Answer 2

Given:-

• A straight line : x = y + 2 touches the circle $4(x^2 + y^2) =r^2$

To Find:-

• Value of r.

Formula used:-

• $\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$

Solution:-

As given, $4(x^2 + y^2) =r^2$

$⟹x^2 + y^2=\dfrac{r²}{2²}$

$⟹x^2 + y^2=(\dfrac{r}{2})²$

And also, x = y + 2 is tangent Then,

• Centre = O(0,0)

• Tangent = x - y - 2 = 0

As we know,

Tangent is always perpendicular to the radius of the circle. So,

Here, x = 0 & y = 0

$\dfrac{r}{2}=\dfrac{।ax_1+by_1+c।}{\sqrt{a^2+b^2}}$

$\dfrac{r}{2}=\dfrac{1(0)-1(0)-2}{\sqrt{1^2+(-1)^2}}$

$\dfrac{r}{2}=\dfrac{2}{\sqrt{2}}$

$r=\dfrac{2×2}{\sqrt{2}}$

$r=\dfrac{4}{\sqrt{2}}$

or,

$r=2\sqrt{2}$

Hence, The value of r is ${\bold{\dfrac{4}{\sqrt{2}}}}$ or 2√2.

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