Question

prove that the line y=2x is a tangent to the circle x^2+y^2-8x-y+5=0​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

Concept: If perpendicular distance of a line from the centre of the circle is equal to radius of the circle, then the line is tangent to the circle.

Here circle is $x^2+y^2-8x-y+5=0$

Comparing it with $x^2+y^2+2gx+2fy+c=0$

we have $g=-4,f=-\frac{1}{2}, c=5$

Thus centre is $(-g,-f)=(4,\frac{1}{2})$ and radius $=\sqrt{g^2+f^2-c}=\sqrt{16+\frac{1}{4}-5}=\sqrt{11+\frac{1}{4}}=\sqrt{\frac{45}{4}}=\frac{3}{2}\sqrt 5$

Now distance of centre from the line $y=2x\Rightarrow 2x-y=0$ is

         $\displaystyle =|\frac{ax_1+b_1+c}{\sqrt{a^2+b^2}}|=|\frac{2(4)-\frac{1}{2}}{\sqrt{2^2+(-1)^2}}|=\frac{\frac{15}{2}}{\sqrt5}=\frac{3}{2}\times \frac{5}{\sqrt 5}=\frac{3}{2}\sqrt 5$

Clearly the distance is same as radius, hence the line is a tangent