Question

The circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4\mathrm{x}+8\mathrm{y}+5$ intersects the line $3\mathrm{x}-4\mathrm{y}=\mathrm{m}$ at two distinct points, if

• A. $-85<\mathrm{m}<–35$
• B. $-35<\mathrm{m}<15$
• C. $15<\mathrm{m}<65$
• D. $35<\mathrm{m}<85$

Answer 1

The correct option is B

$-35<\mathrm{m}<15$

Explanation for the correct option:

Step 1: Finding the radius

Given the equation of circle: ${\mathrm{x}}^2+{\mathrm{y}}^2=4\mathrm{x}+8\mathrm{y}+5$

The line equation: $3\mathrm{x}-4\mathrm{y}=\mathrm{m}$

Since the center of the circle with the general equation ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$is $\left(\frac{-\mathrm{g}}{2},\frac{-\mathrm{f}}{2}\right)$

For the given circle, the Center is $\left(2,4\right)$

and the Radius is given by

$\mathrm{r}=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}$

$$\begin{gathered} =\sqrt{4+16+5} \\ =5 \end{gathered}$$

Step 2: Finding the range of $m$

If the circle is intersecting the line, $3\mathrm{x}-4\mathrm{y}=\mathrm{m}$ at two distinct points,

then the length of the perpendicular from the center should be less than the radius,

$$\begin{gathered} \left|\frac{a^2+b^2-m\left(a+b\right)}{\sqrt{a^2+b^2}}\right| \\ \Rightarrow \left|\frac{6-16-\mathrm{m}}{5}\right|<5 \\ \Rightarrow \left|10+\mathrm{m}\right|<25 \\ \Rightarrow -25<10+\mathrm{m}<25 \\ \Rightarrow -35<\mathrm{m}<15 \end{gathered}$$

Therefore, option (B) is the correct answer.