Question

If the circle $x^2+y^2=a^2$ cuts off a chord of length $2b$ from the line $y=mx+c$,then

• A. $(1-m^2)(a^2-b^2)=c^2$
• B. $(1+m^2)(a^2-b^2)=c^2$
• C. $(1-m^2)(a^2+b^2)=c^2$
• D. $Noneofthese$

Answer 1

Answer: (D)

$Noneofthese$

Consider the given equation of circle

$x^2+y^2=a^2$ …….. (1)

$y=mx+c$ …… (2)

Solving equation (1) $ (2) to, we get,

$x^2+{(mx+c)}^2=a^2$

$\Rightarrow x^2+m^2{x}^2+c^2+2mxc=a^2$

$\Rightarrow (1+m^2)x^2+2mcx+c^2-a^2=0......\left(3\right)$

Compare this equation

$A{x}^2+Bx+C=0$ and we get,

$A=1+m^2$

$B=2mc$

$C=c^2-a^2$

Let $(x_1,x_2)$ be the roots of equation (3)

Then, sum of roots and multiple of roots

$\Rightarrow x_1+x_2=\frac{-B}{A}$ $\Rightarrow x_1.x_2=\frac{C}{A}$

$\Rightarrow x_1+x_2=\frac{-2mc}{1+m^2}$ $\Rightarrow x_1.x_2=\frac{c^2-a^2}{1+m^2}$

Now, given that

Length of chord $=2b$

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

$\Rightarrow {(x_1-x_2)}^2+{(y_1-y_2)}^2=4b^2$

$\Rightarrow {(x_1-x_2)}^2+m^2{(x_1-x_2)}^2=4b^2$

$\Rightarrow {(x_1-x_2)}^2[1+m^2]=4b^2$ since ${(A-B)}^2={(A+B)}^2-4AB$

$\Rightarrow (1+m^2)[{(x_1+x_2)}^2-4x_1{x}_2]=4b^2$

$\Rightarrow (1+m^2)[{\left(\frac{-2mc}{1+m^2}\right)}^2-4\left(\frac{c^2-a^2}{1+m^2}\right)]=4b^2$

$\Rightarrow \frac{4(1+m^2)}{(1+m^2)}[\frac{m^2{c}^2}{(1+m^2)}-\frac{(c^2-a^2)}{1}]=4b^2$

$\Rightarrow m^2{c}^2-(c^2-a^2)(1+m^2)=b^2$

$\Rightarrow m^2{c}^2-[c^2+c^2{m}^2-a^2-a^2{m}^2]=b^2$

$\Rightarrow m^2{c}^2-c^2-c^2{m}^2+a^2+a^2{m}^2=b^2$

$\Rightarrow a^2+a^2{m}^2-c^2-c^2{m}^2=b^2$

$\Rightarrow a^2(1+m^2)-c^2(1+m^2)=b^2$

$\Rightarrow (1+m^2)(a^2-c^2)=b^2$

Hence, this is the answer.

Option (D) is correct.