Question

$B$ and $C$ are points on the circle $x^2+y^2=a^2$
A point $A(b,c)$ lies on that circle such that $AB=AC=d$. The equation to $BC$ is$\cdots \cdots$

• A. $b_X+ay=a^2-d^2$
• B. $b_X+ay=d^2-a^2$
• C. $bx+cy=2a^2-d^2$
• D. $2(bx+cy)=2a^2-d^2$

Answer 1

The correct option is D $2(bx+cy)=2a^2-d^2$

$A(b,c)lieson{x}^2+y^2=a^2$
So, $b^2+c^2=a^2$ ------(1)
Let, $B(x_1,y_1)andc(x_2,y_2)liesoncircle{x}^2+y_2=a^2$
So, $AB=AC=dand{x}_1^2+y_1^2=a^2=y_2^2+x_2^2$
$d=\sqrt{b^2+x_1^2-2b{x}_1+y_1^2+c^2-2y_{1}c}=\sqrt{b^2+x_2^1-2b{x}_2+c^2+y_2^2-2y_{2}c}$ -----(3)
$(x_1^2-x_2^2)-2b(x_1-x_2)+y_1^2-y_2^2-2c(y_1-y_2)=0$
$b(x_1-x_2)+c(y_1-y_2)=0$ -----(2)
So, $e{q}^n$ of of BC is
$y-y_1=(x-x_1)\frac{(y_1-y_2)}{(x_1-x_2)}$
$y-y_1=(x-x_1)\left(\frac{-b}{c}\right)$
$cy+bx=e{y}_1+b{x}_1$
$=\frac{a^2+b^2+c^2-d^2}{2}\left[from\right(3\left)\right]$
$cy+bx=\frac{(2a^2-d^2{)}^2}{2}$
$\Rightarrow 2(bx+cg)=2a^2-d^2$