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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Let $PQ$ be a diameter of the circle $x^{2}+y^{2}=9$. If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line $x+y=2$ respectively,then the maximum value of $\alpha \beta$ is

The perimeter of a certain sector of a circle is equal to the length of the arc of the semicircle. Then,the angle at the centre of the sector in radians is

$P(3,1)$,$Q(6,5)$ and $R(x,y)$ are three points such that the angle $\angle PRQ$ is a right angle and the area of the $\Delta RPQ = 5$. Then the number of such points $R$ is

Find the minimum and maximum distance from the point $(6, 8)$ to the circle $x^2 + y^2 = 4$.

If $(6, -k)$ and $(-3, 2)$ are conjugate points with respect to the circle $x^2 + y^2 + 6x + 4y + 12 = 0$,then $k$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module