Question

Slopes of tangents through (7,1) to the circle $x^2+y^2=25$ satisfy the equation.

• A. $12m^2+7m+12=0$
• B. $12m^2-7m+12=0$
• C. $12m^2+7m-12=0$
• D. $12m^2-7m-12=0$

Answer 1

Answer: (D)

$12m^2-7m-12=0$

We have,

$x^2+y^2=25$

$x^2+y^2=5^2$

Comparing that,

$x^2+y^2=a^2$

Then, centre $(h,k)=(0,0)$ and radius $a=5$units.

We know that the equation of slope

$y=mx\pm a\sqrt{1+m^2}$

$y=mx\pm 5\sqrt{1+m^2}$

Given that,

Tangent passes through $(7,1)$.

$\Rightarrow 1=7m\pm 5\sqrt{1+m^2}$

$\Rightarrow 1-7m=\pm 5\sqrt{1+m^2}$

Squaring both side and we get,

$\Rightarrow {(1-7m)}^2=25(1+m^2)$

$\Rightarrow 1+49m^2-14m=25+25m^2$

$\Rightarrow 24m^2-14m-24=0$

$\Rightarrow 12m^2-7m-12=0$

Hence, this is the answer.