If a line,y=mx+c is a tangent to the circle,( | vedclass

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

Question

(B) The equation of the circle is $x^{2}+y^{2}=1$. The slope of the tangent at $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is found by diff

Vedclass · Accepted Answer

$c^{2}+6c+7=0$

Vedclass · Answer

(B) The equation of the circle is $x^{2}+y^{2}=1$. The slope of the tangent at $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is found by differentiating: $2x+2yy'=0 \Rightarrow y' = -\frac{x}{y}$.At $P$,the slope $m_{L1} = -\frac{1/\sqrt{2}}{1/\sqrt{2}} = -1$.Since the line $y=mx+c$ is perpendicular to $L_{1}$,its slope $m = -\frac{1}{m_{L1}} = -\frac{1}{-1} = 1$.Thus,the line is $y=x+c$,or $x-y+c=0$.This line is a tangent to the circle $(x-3)^{2}+y^{2}=1$,which has center $(3, 0)$ and radius $r=1$.The perpendicular distance from the center $(3, 0)$ to the line $x-y+c=0$ must equal the radius:$\frac{|3-0+c|}{\sqrt{1^{2}+(-1)^{2}}} = 1 \Rightarrow |3+c| = \sqrt{2}$.Squaring both sides: $(3+c)^{2} = 2$ $\Rightarrow 9+6c+c^{2} = 2$ $\Rightarrow c^{2}+6c+7=0$.

If a line,$y=mx+c$ is a tangent to the circle,$(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $L_{1},$ where $L_{1}$ is the tangent to the circle,$x^{2}+y^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

Similar Questions

The equations of the tangents drawn from the origin to the circle $x^2+y^2+2gx+2fy+g^2=0$ are

If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$,then $m_{1}-m_{2}$ is equal to

If $OA$ and $OB$ are the tangents from the origin to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and $C$ is the centre of the circle,the area of the quadrilateral $OACB$ is

For any two nonzero real numbers $a$ and $b$,if the line $\frac{x}{a} + \frac{y}{b} = 1$ is a tangent to the circle $x^2 + y^2 = 1$,then which of the following is true?

The equations of the tangents to the curve $x^2 + y^2 = 10$, which are parallel to the line $2x + y - 1 = 0$, are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module