The equations of tangents to the circle \(x^2+y^2=10\) from the point \((4,-2)\) are
The equations of tangents to the circle \(x^2+y^2=10\) from the point \((4,-2)\) are
\(x+y=2,3 x+2 y=16\)
\(5 x+y=18,3 x-y=4\)
\(3 x+y=10, x-3 y=10\)
\(5 x-y=4, x+y=0\)
Solution
For a tangent, length of perpendicular from origin \(=\) radius.
So we check distance from \((0,0)\) and for any tangent it must be \(\sqrt{10}\).
For \(3 x+y=10\) and \(x-3 y=10\), Distance of \((0,0)\) from line \(3 x+y=10\) is,
\(d=\left|\frac{3 \times 0+0-10}{\sqrt{3^2+1^2}}\right|=\left|\frac{10}{\sqrt{10}}\right|=\sqrt{10}\)
Also from \(x-3 y=10\),
\(d=\left|\frac{0-3 \times 0-10}{\sqrt{3^2+1^2}}\right|=\sqrt{10}\)
So, option (c) is correct.