If 5x - 12y + 10 = 0 and 12y - 5x + 16 = 0 ar | vedclass

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

Question

(A) Given the two tangent lines to the circle are: $5x - 12y + 10 = 0$ $\dots$ $(i)$ $-5x + 12y + 16 = 0$ $\dots$ $(ii)$ Since the slopes of both line

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the two tangent lines to the circle are: $5x - 12y + 10 = 0$ $\dots$ $(i)$ $-5x + 12y + 16 = 0$ $\dots$ $(ii)$ Since the slopes of both lines are equal to $\frac{5}{12}$,the lines are parallel. The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 5, B = -12, C_1 = 10, C_2 = 16$. The distance between the tangents is the diameter of the circle: $d = \frac{|10 - 16|}{\sqrt{5^2 + (-12)^2}} = \frac{|-6|}{\sqrt{25 + 144}} = \frac{6}{13}$. Wait,re-evaluating the constant terms: Equation $(i): 5x - 12y + 10 = 0$ Equation $(ii): -5x + 12y + 16 = 0 \implies 5x - 12y - 16 = 0$ Distance $d = \frac{|10 - (-16)|}{\sqrt{5^2 + (-12)^2}} = \frac{26}{13} = 2$. Since the distance between parallel tangents is the diameter,$2r = 2$. Therefore,the radius $r = 1$.

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

Similar Questions

The equation of a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ passing through the origin is:

If a circle is inscribed in an equilateral triangle of side $a$,then the area of any square (in sq. units) inscribed in this circle is

The set of values of $k$ for which the circle $C : 4x^{2} + 4y^{2} - 12x + 8y + k = 0$ lies inside the fourth quadrant and the point $(1, -1/3)$ lies on or inside the circle $C$ is

The number of possible common tangents that can be drawn to the circles $x^2+y^2+4x-6y-3=0$ and $x^2+y^2+4x-2y+1=0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module