The equation of the tangent to the circle at | vedclass

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Question

(A) The center of the circle is the point of intersection of the lines $x - y = 1$ and $2x + y = 3$.Adding the two equations: $(x - y) + (2x + y) = 1

Vedclass · Accepted Answer

$x + 4y + 3 = 0$

Vedclass · Answer

(A) The center of the circle is the point of intersection of the lines $x - y = 1$ and $2x + y = 3$.Adding the two equations: $(x - y) + (2x + y) = 1 + 3 \implies 3x = 4 \implies x = \frac{4}{3}$.Substituting $x = \frac{4}{3}$ into $x - y = 1$: $\frac{4}{3} - y = 1 \implies y = \frac{4}{3} - 1 = \frac{1}{3}$.So,the center $O$ is $\left(\frac{4}{3}, \frac{1}{3}\right)$.The point of tangency is $P(1, -1)$.The slope of the radius $OP$ is $m_{OP} = \frac{\frac{1}{3} - (-1)}{\frac{4}{3} - 1} = \frac{\frac{4}{3}}{\frac{1}{3}} = 4$.Since the tangent is perpendicular to the radius at the point of contact,the slope of the tangent $m_t = -\frac{1}{m_{OP}} = -\frac{1}{4}$.The equation of the tangent passing through $(1, -1)$ with slope $-\frac{1}{4}$ is:$y - (-1) = -\frac{1}{4}(x - 1)$$4(y + 1) = -(x - 1)$$4y + 4 = -x + 1$$x + 4y + 3 = 0$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

The equation of the tangent to the circle at the point $(1, -1)$,whose center is the point of intersection of the straight lines $x - y = 1$ and $2x + y = 3$,is:

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