The tangent to the circle \(x^2+y^2 = 10\) at the point \((3,1)\) touches the circle \(x^2+y^2-2\sqrt{10}\,x-20y+k = 0\), then the value of \(k\) is...
Show Hint
The tangent is a tangent to the second circle too, so its distance from that centre equals its radius.
Step 1: Write the second circle in standard form:
$(x - \sqrt{10})^2 + (y - 10)^2 = 10 + 100 - k$.
Step 2: Tangency condition:
The perpendicular distance from $(\sqrt{10}, 10)$ to $3x + y - 10 = 0$ is $3$, so $r^2 = 9$.
$110 - k = 9$ gives $k = 101$. Any other value of $k$ gives $r^2
eq 9$, so the circle would cut the line or miss it.
Final Answer:
$k = 101$, option (D).
\[ \boxed{101} \]