If x-2y=0 is a tangent drawn at a point P on | vedclass

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

Question

(B) The equation of the circle is $x^2+y^2-6x+2y+c=0$. The center of the circle is $C = (3, -1)$.Since $x-2y=0$ is a tangent at point $P$,the radius $

Vedclass · Accepted Answer

$2\sqrt{5}$

Vedclass · Answer

(B) The equation of the circle is $x^2+y^2-6x+2y+c=0$. The center of the circle is $C = (3, -1)$.Since $x-2y=0$ is a tangent at point $P$,the radius $CP$ is perpendicular to the tangent line.The slope of the tangent $x-2y=0$ is $m_1 = 1/2$. Thus,the slope of the normal $CP$ is $m_2 = -2$.The equation of the normal line passing through $C(3, -1)$ with slope $-2$ is $y - (-1) = -2(x - 3)$,which simplifies to $y+1 = -2x+6$,or $2x+y-5=0$.Point $P$ is the intersection of the tangent $x-2y=0$ and the normal $2x+y-5=0$.Solving these: $x=2y$,so $2(2y)+y-5=0 \implies 5y=5 \implies y=1$. Then $x=2(1)=2$. So $P = (2, 1)$.The distance between $P(2, 1)$ and the point $(6, 3)$ is given by the distance formula:$d = \sqrt{(6-2)^2 + (3-1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.

If $x-2y=0$ is a tangent drawn at a point $P$ on the circle $x^2+y^2-6x+2y+c=0$,then the distance of the point $(6,3)$ from $P$ is

Similar Questions

If the tangent to the conic $y - 6 = x^2$ at $(2, 10)$ touches the circle $x^2 + y^2 + 8x - 2y = k$ (for some fixed $k$) at a point $(\alpha, \beta)$,then $(\alpha, \beta)$ is

If $2x - 3y + 5 = 0$ and $4x - 5y + 7 = 0$ are the equations of the normals drawn to a circle and $(2, 5)$ is a point on the given circle,then the radius of the circle is

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

Two common tangents to the circle ${x^2} + {y^2} = 2{a^2}$ and the parabola ${y^2} = 8ax$ are

The equations of the tangents to the circle $x^2 + y^2 = 50$ at the points where the line $x + 7 = 0$ meets it,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module