If the line y = mx + C is a tangent to the ci | vedclass

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

Question

(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.We know that the line $y = mx + C$ is

Vedclass · Accepted Answer

$\pm \frac{1}{4} \sqrt{C^2 - 16}$

Vedclass · Answer

(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.We know that the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = a^2$ if and only if $C^2 = a^2(1 + m^2)$.Substituting $a^2 = 16$ into the condition,we get $C^2 = 16(1 + m^2)$.Dividing by $16$,we have $\frac{C^2}{16} = 1 + m^2$.Rearranging for $m^2$,we get $m^2 = \frac{C^2}{16} - 1 = \frac{C^2 - 16}{16}$.Taking the square root on both sides,we get $m = \pm \frac{1}{4} \sqrt{C^2 - 16}$.Thus,the correct option is $B$.

If the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = 16$,then $m =$

Similar Questions

If the length of the tangent drawn from the point $(5, 3)$ to the circle $x^2 + y^2 + ky + 17 = 0$ is $7$,then $k = \dots$

If the equation of the common tangent at the point $(1, -1)$ to the two circles,each of radius $13$,is $12x + 5y - 7 = 0$,then the centers of the two circles are

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$,then its radius is

The equation of the tangents to the circle $x^2 + y^2 + 4x - 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

The equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$. The line joining the points $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then $17\beta - \alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module