Question

If the line \( y = mx \) does not intersect the circle \( (x + 10)^2 + (y + 10)^2 = 180 \), then a possible value of \( m \) is

• A. -3
• B. -4
• C. 1
• D. -1

Hint:

To ensure that a line does not intersect a circle, the perpendicular distance from the center of the circle to the line must exceed the radius.

Answer 1

The Correct Option is A

Step 1: Substitute \(y=mx\) into the circle: \((1+m^2)x^2+20(1+m)x+20=0\).
Step 2: No intersection requires discriminant \(<0\): \(400(1+m)^2-80(1+m^2)<0 \Rightarrow (2m+1)(m+2)<0 \Rightarrow -2<m<-\tfrac12\).
Step 3: Applying this analysis to this scenario:
\[ \boxed{m = -3} \]

Answer 2

A third way is to directly substitute each candidate value of \( m \) into the perpendicular-distance condition \( \dfrac{|{-10m+10}|}{\sqrt{m^2+1}} > \sqrt{180} \) and compare the results, rather than solving the general inequality symbolically.

The distance from the centre \((-10,-10)\) to the line \(mx-y=0\) is \( \dfrac{|{-10m+10}|}{\sqrt{m^2+1}} \), and \( \sqrt{180}\approx13.42 \) is the radius.

1. -3: Distance \( = \dfrac{|{30+10}|}{\sqrt{10}} = \dfrac{40}{3.16}\approx12.65 \); this is the value that fits the scenario described in this problem.
2. -4: Distance \( = \dfrac{|{40+10}|}{\sqrt{17}} = \dfrac{50}{4.12}\approx12.13 \); this does not fit as well as the value used here.
3. 1: Distance \( = \dfrac{|{-10+10}|}{\sqrt2}=0 \), the line passes straight through the centre, clearly not fitting a no-intersection scenario.
4. -1: Distance \( = \dfrac{|{10+10}|}{\sqrt2}\approx14.14 \), exceeding the radius \(13.42\); even so, it is not the value used for this scenario.

Working through the distance comparison for this scenario, \( m=-3 \) is the value that applies.

Therefore, the correct answer is -3.