A circle with centre (a, b) passes through th | vedclass

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

Question

(B) The center of the circle is $C = (a, b)$ and it passes through the origin $O = (0, 0)$.The radius $OC$ connects the origin to the center. The slop

Vedclass · Accepted Answer

$ax + by = 0$

Vedclass · Answer

(B) The center of the circle is $C = (a, b)$ and it passes through the origin $O = (0, 0)$.The radius $OC$ connects the origin to the center. The slope of the radius $OC$ is $m_{radius} = \frac{b - 0}{a - 0} = \frac{b}{a}$.The tangent at the origin is perpendicular to the radius $OC$ at the point of contact $(0, 0)$.Therefore,the slope of the tangent $m_{tangent}$ is the negative reciprocal of the slope of the radius:$m_{tangent} = -\frac{1}{m_{radius}} = -\frac{a}{b}$.Using the point-slope form of a line passing through $(0, 0)$ with slope $m = -\frac{a}{b}$:$y - 0 = -\frac{a}{b}(x - 0)$$y = -\frac{a}{b}x$$by = -ax$$ax + by = 0$.

$A$ circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

Similar Questions

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$,then a value of $\alpha$ is

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at $P(-4,0)$. Then the area of quadrilateral $PAOB$,where $O$ is the origin,is

The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

If the tangents $x+y+k=0$ and $x+ay+b=0$ drawn to the circle $S \equiv x^2+y^2+2x-2y+1=0$ are perpendicular to each other and $k, b$ are both greater than $1$,then $b-k=$

The equation of the tangent to the circle $x^2+y^2=25$ at $(-3, 4)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module