Question

If the lines $L_1$ and $L_2$ are tangents to $4x^2-4x-24y+49=0$ and are normals for $x^2+y^2=72$, then find the slopes of $L_1$ and $L_2$

Answer 1

We know normal of a circle passes through the centre of it.

$Hence{L}_1{L}_{2}willpassthroughcentreof{x}^2+y^2=72i.e.(0,0).Eq.of{L}_1\left(slope{m}_1\right)and{L}_2\left(slope{m}_2\right).L_1:y-0=m_1(x-0)y=m_{1}x.L_2:y=m_{2}x.\therefore Onputtingy=mxinequation4x^2-4x-24y+49=0,therootsmustbesuchthatthediscriminant=0.\Rightarrow 4x^2-4x-24\left(mx\right)+49=0.\Rightarrow 4x^2+x(-4-24m)+49=0.\Rightarrow (4+24m{)}^2-4(4\left)\right(49)=0\Rightarrow (\because D=b^2-4ac=0)\Rightarrow (4+24m{)}^2=4\times 4\times 49.\Rightarrow 4+24m=\pm (2\times 2\times 7)\Rightarrow 4+24m=28\Rightarrow 24m=24\Rightarrow m=1and4+24m=-28.\Rightarrow 24m=-32\to m=\frac{-32}{24}=\frac{-4}{3}.\therefore L_1:y=x{L}_2:y=\frac{-4}{3}x\therefore Slopesare1and\frac{-4}{3}respectively.$