1. Find the centre of the circle

Given parabola: $y^2=4x+16=4(x+4)$ This is of the form $y^2=4a(x-h)$ with vertex at $(-4,0)$ and $a=1$.

So its focus is: $(h+a,0)=(-4+1,0)=(-3,0)$

Hence the circle $C$ has centre $(-3,0)$ and radius $5$.

Therefore its equation is $(x+3)^2+y^2=25$

2. Find the intersection point of the two lines

The lines are: $3x-y=0$ $x+\lambda y=4$

From the first line, $y=3x$

Substitute into the second line: $x+\lambda(3x)=4$ $x(1+3\lambda)=4$ $x=\frac{4}{1+3\lambda}$

Then $y=3x=\frac{12}{1+3\lambda}$

So the intersection point is $\left(\frac{4}{1+3\lambda},\frac{12}{1+3\lambda}\right)$

3. Impose the condition that this point lies on the circle

Substitute into $(x+3)^2+y^2=25$

We get $\left(\frac{4}{1+3\lambda}+3\right)^2+\left(\frac{12}{1+3\lambda}\right)^2=25$

Let $t=1+3\lambda$ Then $x=\frac{4}{t}$ and $y=\frac{12}{t}$.

So $\left(\frac{4}{t}+3\right)^2+\left(\frac{12}{t}\right)^2=25$

Expand: $\frac{(4+3t)^2}{t^2}+\frac{144}{t^2}=25$ $\frac{16+24t+9t^2+144}{t^2}=25$ $\frac{160+24t+9t^2}{t^2}=25$

Multiply by $t^2$: $160+24t+9t^2=25t^2$ $16t^2-24t-160=0$ $2t^2-3t-20=0$

Solve: $2t^2-3t-20=(2t+5)(t-4)=0$

Hence $t=4 \quad \text{or} \quad t=-\frac52$

Now recall $t=1+3\lambda$

So:

• If $t=4$, $1+3\lambda=4 \Rightarrow 3\lambda=3 \Rightarrow \lambda=1$
• If $t=-\frac52$, $1+3\lambda=-\frac52 \Rightarrow 3\lambda=-\frac72 \Rightarrow \lambda=-\frac76$

Thus $\lambda_1=-\frac76, \qquad \lambda_2=1$

4. Compute the required value

$12\lambda_1+29\lambda_2=12\left(-\frac76\right)+29(1)$ $=-14+29=15$

5. Comparison with stored answer

Derived answer is: $15$ This matches the stored correct answer.