Question

find m for which the line 3x-my+2=0 intersects the circle
${x}^{2} + {y}^{2} - 4x + 6y - 3= 0$
at 2 coincident points. (tangent to the circle)

plz ans with explanation and don't spam or will report you​

Answer 1

$\hugem=\dfrac{20}{7}\text{ or }m=4$

$\huge\text{\underline{\underline{Explanation}}}$

$\large\boxed{\text{Geometry}}$

We know that the tangent and the circle are apart by radius.

$\cdots\longrightarrow d=r$

$\large\boxed{\text{Standard form}}$

$\cdots\longrightarrow(x^{2}-4x+4)+(y^{2}+6y+9)-4-9-3=0$

$\cdots\longrightarrow(x-2)^{2}+(y+3)^{2}=4^{2}$

$\large\boxed{\text{Properties of the circle}}$

$\text{ \bullet \text{The circle is centered at }(2,-3) .}$

$\text{ \bullet \text{The radius of the circle is }4 .}$

$\large\boxed{\text{Distance formula}}$

$\cdots\longrightarrow d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

$\text{ \bullet\ a, b, c are the coefficients.}$

$\text{ \bullet\ (x_{1}.y_{1}) is the coordinate.}$

$\large\boxed{\text{Solving for }m}$

Now since we know $a=3$, $b=-m$, $c=2$, $d=4$ and $(2,-3)$, we want the equation for $m$.

$\cdots\longrightarrow 4=\dfrac{|6+3m+2|}{\sqrt{9+m^{2}}}$

This is the wanted equation. Let's break this down.

$\cdots\longrightarrow 16(9+m^{2})=(6+3m+2)^{2}$

$\cdots\longrightarrow 144+16m^{2}=9m^{2}+48m+64$

$\cdots\longrightarrow7m^{2}-48m+80=0$

$\cdots\longrightarrow(7m-20)(m-4)=0$

$\cdots\longrightarrow m=\dfrac{20}{7}\text{ or }m=4$

$\huge\text{\underline{\underline{Final answer}}}$

Hence, -

$\text{ \cdots\longrightarrow\boxed{m=\dfrac{20}{7}\text{ or }m=4} }$