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Find the equation of the circle whose center is on the line $4x-3y=0$ and tangent to the lines $4x-3y-25=0$ and $3x-4y+32=0$.

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  • $\begingroup$ Do you have any idea? For example, wouldn't we need to have a point on the line the center is on which is at the same distance from both other two lines...? $\endgroup$ – DonAntonio Jan 6 '17 at 13:13
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First note that any point on the line $4x-3y=0$ can be written as $(3k,4k)$ for some $k$. Now as we know that tangent is perpendicular to the line joining centre and point of tangency, we have the radius equal to the perpendicular distance of centre from the tangent line. With that information and the fact that perpendicular distance of point $(x_0,y_0)$ from line $ax+by+c=0$ is $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$, we have radius equal to $5$ as the perpendicular distance of any point of the form $(3k,4k)$ from $4x-3y-25=0$ is $5$. Now using $r=5$ and the fact the distance of centre from $3x-4y+32=0$ is equal to radius we can get solution for $k$, one of which is $k=1$. So our centre can be $(3,4)$ and desired equation of the circle is $(x-3)^2+(y-4)^2=5^2$

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