If the foot of the normal from the point (4,3 | vedclass

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

Question

(C) The normal at any point on a circle always passes through its center. The normal passes through $(4,3)$ and $(2,1)$.The equation of this normal is

Vedclass · Accepted Answer

$x^2+y^2-2x-1=0$

Vedclass · Answer

(C) The normal at any point on a circle always passes through its center. The normal passes through $(4,3)$ and $(2,1)$.The equation of this normal is $y-1 = \frac{3-1}{4-2}(x-2)$ $\Rightarrow y-1 = 1(x-2)$ $\Rightarrow y = x-1 \dots(1)$.The center of the circle lies on this normal. We are also given that $2x-y-2=0$ is a diameter,so the center also lies on this line $\dots(2)$.Solving $(1)$ and $(2)$ by substituting $y = x-1$ into $2x-(x-1)-2=0$,we get $x-1=0 \Rightarrow x=1$. Then $y=0$. Thus,the center is $(1,0)$.The radius $r$ is the distance between the center $(1,0)$ and the point $(2,1)$ on the circle: $r = \sqrt{(2-1)^2 + (1-0)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.The equation of the circle with center $(1,0)$ and radius $\sqrt{2}$ is $(x-1)^2 + (y-0)^2 = (\sqrt{2})^2$ $\Rightarrow x^2-2x+1+y^2=2$ $\Rightarrow x^2+y^2-2x-1=0$.

If the foot of the normal from the point $(4,3)$ to a circle is $(2,1)$ and $2x-y-2=0$ is a diameter of the circle,then the equation of the circle is

Similar Questions

If the shortest distance from $(2,-14)$ to the circle $x^2+y^2+6x+4y-12=0$ is $d$ and the length of the tangent drawn from the same point to the circle is $l$,then $\sqrt{d+l}=$

If two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points,then

The length of the intercept on the line $4x - 3y - 10 = 0$ by the circle $x^2 + y^2 - 2x + 4y - 20 = 0$ is

Let $l > 0$ be a real number,$C$ denote a circle with circumference $l$,and $T$ denote a triangle with perimeter $l$. Then:

Statement $(A):$ The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is $4$.
Reason $(R):$ For two circles with centers $C_1, C_2$ and radii $r_1, r_2$,if $|C_1C_2| > r_1 + r_2$,then the circles have $4$ common tangents.

Vedclass Test Series

Exam Paper Generator

Online Exam Module