The normal to the circle given by x^2+y^2-6x+ | vedclass

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

Question

(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightar

Vedclass · Accepted Answer

$(-2,-16)$

Vedclass · Answer

(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightarrow g = -3$ and $2f = 8 \Rightarrow f = 4$. The centre of the circle is $(-g, -f) = (3, -4)$. We know that the normal to a circle at any point always passes through its centre. Let the point where the normal meets the circle again be $(x_1, y_1)$. Since the centre $(3, -4)$ is the midpoint of the chord connecting $(8, 8)$ and $(x_1, y_1)$,we have: $\frac{x_1+8}{2} = 3$ $\Rightarrow x_1+8 = 6$ $\Rightarrow x_1 = -2$ $\frac{y_1+8}{2} = -4$ $\Rightarrow y_1+8 = -8$ $\Rightarrow y_1 = -16$ Thus,the required point is $(-2, -16)$.

The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

Similar Questions

The equation of the circle which touches the circle ${x^2 + y^2 - 6x + 6y + 17 = 0}$ externally and to which the lines ${x^2 - 3xy - 3x + 9y = 0}$ are normals,is

The equations of the tangents to the circle $x^{2}+y^{2}-6x+4y=12$,which are parallel to the straight line $4x+3y+5=0$,are

The angle between the two tangents drawn from the origin to the circle $x^2+y^2-14x+2y+25=0$ is (in $^{\circ}$)

If the line $4x - 3y + p = 0$ $(p + 3 > 0)$ touches the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ at the point $(h, k)$,then $h - 2k = . . . . . .$

Let $AB$ be a chord of length $12$ of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$. If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$,then five times the distance of point $P$ from chord $AB$ is equal to $.......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module