If the circle (x - h)^2 + (y - k)^2 = r^2 tou | vedclass

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Question

(B) The circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$.This implies that the normal to the curve $y = x^2

Vedclass · Accepted Answer

$h + 2k = 5$

Vedclass · Answer

(B) The circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$.This implies that the normal to the curve $y = x^2 + 1$ at $(1, 2)$ must pass through the center of the circle $(h, k)$.The derivative of the curve is $dy/dx = 2x$. At $x = 1$,the slope of the tangent is $m_t = 2(1) = 2$.The slope of the normal at $(1, 2)$ is $m_n = -1/m_t = -1/2$.The equation of the normal line passing through $(1, 2)$ with slope $-1/2$ is $(y - 2) = -1/2(x - 1)$,which simplifies to $2y - 4 = -x + 1$,or $x + 2y = 5$.Since the center $(h, k)$ lies on this normal line,we have $h + 2k = 5$.

If the circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$,then the possible locations of the points $(h, k)$ are given by

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