If a circle with centre at (-1, 1) touches th | vedclass

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Question

(A) The point of contact is the foot of the perpendicular from the centre $(-1, 1)$ to the line $x + 2y + 4 = 0$. Let the point of contact be $(h, k)$

Vedclass · Accepted Answer

$(-2, -1)$

Vedclass · Answer

(A) The point of contact is the foot of the perpendicular from the centre $(-1, 1)$ to the line $x + 2y + 4 = 0$. Let the point of contact be $(h, k)$. The line passing through the centre and perpendicular to the given line has a slope of $2$ (since the slope of $x + 2y + 4 = 0$ is $-1/2$). The equation of this perpendicular line is $y - 1 = 2(x + 1)$,which simplifies to $y = 2x + 3$. Substituting $y = 2x + 3$ into the given line equation $x + 2(2x + 3) + 4 = 0$: $x + 4x + 6 + 4 = 0$ $5x + 10 = 0$ $x = -2$. Substituting $x = -2$ into $y = 2x + 3$: $y = 2(-2) + 3 = -1$. Thus,the point of contact is $(-2, -1)$.

If a circle with centre at $(-1, 1)$ touches the line $x + 2y + 4 = 0$,then the coordinates of the point of contact are:

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