If the equation of the tangent to the circle | vedclass

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

Question

(A) The given equation of the circle is $x^2 + y^2 - 2x + 6y - 6 = 0$.Completing the square,we get $(x - 1)^2 + (y + 3)^2 = 6 + 1 + 9 = 16 = 4^2$.Thus

Vedclass · Accepted Answer

$5, -35$

Vedclass · Answer

(A) The given equation of the circle is $x^2 + y^2 - 2x + 6y - 6 = 0$.Completing the square,we get $(x - 1)^2 + (y + 3)^2 = 6 + 1 + 9 = 16 = 4^2$.Thus,the center of the circle is $(1, -3)$ and the radius $r = 4$.The equation of the tangent parallel to $3x - 4y + 7 = 0$ is of the form $3x - 4y + k = 0$.The perpendicular distance from the center $(1, -3)$ to the tangent line $3x - 4y + k = 0$ must be equal to the radius $r = 4$.Using the distance formula $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = r$,we have $\frac{|3(1) - 4(-3) + k|}{\sqrt{3^2 + (-4)^2}} = 4$.$\frac{|3 + 12 + k|}{5} = 4$.$|15 + k| = 20$.This gives $15 + k = 20$ or $15 + k = -20$.Therefore,$k = 5$ or $k = -35$.

If the equation of the tangent to the circle $x^2 + y^2 - 2x + 6y - 6 = 0$ parallel to $3x - 4y + 7 = 0$ is $3x - 4y + k = 0$,then the values of $k$ are

Similar Questions

The equation of a circle which has a tangent $3x + 4y = 6$ and two normals given by $(x - 1)(y - 2) = 0$ is

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

The equation of the normal to the circle $x^2 + y^2 = 9$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ is

The equation of the normal at the point $(4, -1)$ of the circle $x^2 + y^2 - 40x + 10y = 153$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module