Question:
A line $y = mx + 1$ intersects the circle $(x ? 3)^2 + (y + 2)^2 = 25$ at the points $P$ and $Q$ If the midpoint of the line segment $PQ$ has $x$-coordinate $-\frac{3}{5}$, then which one of the following options is correct?
A line $y = mx + 1$ intersects the circle $(x ? 3)^2 + (y + 2)^2 = 25$ at the points $P$ and $Q$ If the midpoint of the line segment $PQ$ has $x$-coordinate $-\frac{3}{5}$, then which one of the following options is correct?
Updated On: Jun 14, 2022
- $-3\le m
- $2\le m<4$
- $4\le m<6$
- $6\le m<8$
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The Correct Option is B
Solution and Explanation
The mid-point of chord PQ can be considered as $A\left(-\frac{3}{5}, -\frac{3}{5}m+1\right)$
$\because AO \bot PQ
Slope of AO ? Slope of PQ = -1$
$\Rightarrow\left(\frac{1-\frac{3}{5}m+2}{-\frac{3}{5}-3}\right).m=-1$
$\Rightarrow \left(3-\frac{3}{5}m\right)m=\frac{18}{5}$
$\Rightarrow m^{2}-5m+6=0$
$\Rightarrow m=2$ or $3$
$\because AO \bot PQ
Slope of AO ? Slope of PQ = -1$
$\Rightarrow\left(\frac{1-\frac{3}{5}m+2}{-\frac{3}{5}-3}\right).m=-1$
$\Rightarrow \left(3-\frac{3}{5}m\right)m=\frac{18}{5}$
$\Rightarrow m^{2}-5m+6=0$
$\Rightarrow m=2$ or $3$
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Concepts Used:
The Slope of a Line
A slope of a line is the conversion in y coordinate w.r.t. the conversion in x coordinate.
The net change in the y-coordinate is demonstrated by Δy and the net change in the x-coordinate is demonstrated by Δx.
Hence, the change in y-coordinate w.r.t. the change in x-coordinate is given by,
\(m = \frac{\text{change in y}}{\text{change in x}} = \frac{Δy}{Δx}\)
Where, “m” is the slope of a line.
The slope of the line can also be shown by
\(tan θ = \frac{Δy}{Δx}\)
Read More: Slope Formula
The slope of a Line Equation:
The equation for the slope of a line and the points are known to be a point-slope form of the equation of a straight line is given by:
\(y-y_1=m(x-x_1)\)
As long as the slope-intercept form the equation of the line is given by:
\(y = mx + b\)
Where, b is the y-intercept.