Question
Question: The shortest distance from the line \[3x\text{ + 4y = 25}\] to the circle \[{{x}^{2}}\text{ + }{{\te...
The shortest distance from the line 3x + 4y = 25 to the circle x2 + y2 = 6x - 8y is equal to
& A.\dfrac{1}{2} \\\ & B.\dfrac{7}{5} \\\ & C.\dfrac{11}{5} \\\ & D.\dfrac{1}{3} \\\ \end{aligned}$$Solution
As we know the equation of circle so transform it into (x - x1)2 + (y - y1)2 = r2 where (x1, y1) center’s point and r is radius. Then to find shortest distance, first find distance between center of circle and the given line using formula
\dfrac{\left| Aa+Bb+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$$$$$$
Where equation of line is \text{Ax+By+C}=0$$ and point is (a, b) and then subtract it with the given radius r.
Complete step-by-step solution:
In the question, equation of a line 3x + 4y = 25 and the circle x2 + y2 = 6x - 8y is given and we have to find shortest distance between them.
So we can represent it by the diagram,
Let AB be the shortest distance between line and circle. Let O be the center of the circle so OA can be represented as the radius of the circle.
So we can represent it by the diagram,
The circle and line represented in the diagram have equations 3x + 4y = 25 and x2 + y2 = 6x - 8y respectively.
The given equation of circle is,
x2 + y2 = 6x - 8y
Which we can write it as,
x2 + y2 - 6x + 8y = 0
Now we will transform it into the form (x - x1)2 + (y - y1)2 = r2 where (x1, y1) point of center and r is is radius. Here we add and subtract the square of half of the coefficient of x and y.
So we can write the equation of circles,