Given the circle C with the equation x^2+y^2- | vedclass

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(A) The given equation of the circle is $x^2+y^2-2x+10y-38=0$.The center of the circle is $(1, -5)$.$(A)$ The equation of the polar of $(x_1, y_1)$ wi

Vedclass · Accepted Answer

$A-III, B-IV, C-I, D-II$

Vedclass · Answer

(A) The given equation of the circle is $x^2+y^2-2x+10y-38=0$.The center of the circle is $(1, -5)$.$(A)$ The equation of the polar of $(x_1, y_1)$ with respect to $x^2+y^2+2gx+2fy+c=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.For $(4, 3)$,the equation is $x(4)+y(3)-1(x+4)+5(y+3)-38=0$.$\Rightarrow 4x+3y-x-4+5y+15-38=0$ $\Rightarrow 3x+8y-27=0$ $\Rightarrow 3x+8y=27$ $(III)$.$(B)$ The equation of the tangent at $(x_1, y_1)$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.For $(9, -5)$,the equation is $x(9)+y(-5)-1(x+9)+5(y-5)-38=0$.$\Rightarrow 9x-5y-x-9+5y-25-38=0$ $\Rightarrow 8x-72=0$ $\Rightarrow x=9$ $(IV)$.$(C)$ The normal at any point on a circle passes through the center $(1, -5)$.The normal at $(-7, -5)$ passes through $(-7, -5)$ and $(1, -5)$.Since the $y$-coordinates are the same,the equation is $y = -5$,or $y+5=0$ $(I)$.$(D)$ The diameter passes through the center $(1, -5)$ and the point $(1, 3)$.Since the $x$-coordinates are the same,the equation is $x=1$ $(II)$.Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

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