Q90 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Mathematics

The equations of tangents to the circle \(x^2+y^2=10\) from the point \((4,-2)\) are

A
\(x+y=2,3 x+2 y=16\)
B
\(5 x+y=18,3 x-y=4\)
Correct answer
C
\(3 x+y=10, x-3 y=10\)
D
\(5 x-y=4, x+y=0\)

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.

Hence, the correct answer is option: 'C'.

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