Question
The length of the diameter of the circle which has the lines $12 x+5 y+52=0$ and $12 x+5 y=0$ as its tangents is(a) 5(b) 4(c) 42(d) 1
Step 1
We can see that these two lines are parallel because they have the same slope, which is $-\frac{12}{5}$. Show more…
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The length of the tangent from any point on the circle $x^{2}+y^{2}-8 x+5 y-4=0$ to the circle $x^{2}+y^{2}-8 x+5 y=0$ is (a) 1 (b) 2 (c) 4 (d) 6
Let the tangents drawn from the origin to the circle, $x^{2}+y^{2}-8 x-4 y+$ $16=0$ touch it at the points $A$ and $B$. The $(A B)^{2}$ is equal to: (a) $\frac{52}{5}$ (b) $\frac{56}{5}$ (c) $\frac{64}{5}$ (d) $\frac{32}{5}$
A tangent to the circle $x^{2}+y^{2}=1$ through the point $(0,5)$ cuts the circle $x^{2}+y^{2}=4$ at $A$ and $B$. The tangents for the circle $x^{2}+y^{2}=4$ at $A$ and $B$ meet at $C$. The coordinates of $C$ are (A) $\left(\frac{8 \sqrt{6}}{5}, \frac{4}{5}\right)$ (B) $\left(-\frac{8 \sqrt{6}}{5}, \frac{4}{5}\right)$ (C) $\left(\frac{8 \sqrt{6}}{5},-\frac{4}{5}\right)$ (D) $\left(-\frac{8 \sqrt{6}}{5},-\frac{4}{5}\right)$
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