Question

Find the equation of tangent to circle $x^2+y^2=64$ which passes through the point $\left(4,7\right)$ .

Answer 1

given equation of circle is $x^2+y^2=64$

As we know that

The equation of tangent to the circle whose slope is $m$ is $y=mx\pm a\sqrt{1+m^2}$

$⟹y=mx\pm 8\sqrt{1+m^2}$

$(4,7)$ lies on $y=mx\pm 8\sqrt{1+m^2}$

$⟹(7-4m{)}^2=64(1+m^2)$

$⟹49+16m^2-56m=64+64m^2$

$⟹48m^2+56m+15=0$

$⟹48m^2+36m+20m+15=0⟹12m(4m+3)+5(4m+3)=0$

$⟹m=-\frac{5}{12},-\frac{3}{4}$

The equations of tangents will be $y=-\frac{5}{12}x\pm 8\times \frac{13}{12},y=-\frac{3}{4}x\pm 8\times \frac{5}{4}⟹5x+12y=\pm 104,3x+4y=\pm 40$