JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 3)
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 $$ - {3 \over 5}$$, then which one of the following options is correct?
6 $$ \le $$ m < 8
$$ - $$3 $$ \le $$ m < $$ - $$1
4 $$ \le $$ m < 6
2 $$ \le $$ m < 4
Explanation
It is given that points P and Q are intersecting points of circle $${(x - 3)^2} + {(y + 2)^2}$$ = 25 .....(i)
Line y = mx + 1 .....(ii)
And, the mid-point of PQ is A having x-coordinate $$ - {3 \over 5}$$
so y-coordinate is $$1 - {3 \over 5}$$ m.
So, $$A\left( { - {3 \over 5},1 - {3 \over 5}m} \right)$$
From the figure,
$$ \because $$ AC $$ \bot $$ PQ
$$ \Rightarrow $$ (slope of AC) $$ \times $$ (slope of PQ) = $$ - $$1
$$ \Rightarrow $$ $$\left( {{{ - 2 - 1 + {3 \over 5}m} \over {3 + {3 \over 5}}}} \right) \times m = - 1$$
$$ \Rightarrow {{(3/5)m - 3} \over {18/5}}m = - 1$$
$$ \Rightarrow \left( {{{3m - 15} \over {18}}} \right)m = - 1$$
$$ \Rightarrow 3{m^2} - 15m + 18 = 0$$
$$ \Rightarrow {m^2} - 5m + 6 = 0$$
$$ \Rightarrow $$ m = 2 or 3
Line y = mx + 1 .....(ii)
And, the mid-point of PQ is A having x-coordinate $$ - {3 \over 5}$$
so y-coordinate is $$1 - {3 \over 5}$$ m.
So, $$A\left( { - {3 \over 5},1 - {3 \over 5}m} \right)$$
From the figure,
$$ \because $$ AC $$ \bot $$ PQ
$$ \Rightarrow $$ (slope of AC) $$ \times $$ (slope of PQ) = $$ - $$1
$$ \Rightarrow $$ $$\left( {{{ - 2 - 1 + {3 \over 5}m} \over {3 + {3 \over 5}}}} \right) \times m = - 1$$
$$ \Rightarrow {{(3/5)m - 3} \over {18/5}}m = - 1$$
$$ \Rightarrow \left( {{{3m - 15} \over {18}}} \right)m = - 1$$
$$ \Rightarrow 3{m^2} - 15m + 18 = 0$$
$$ \Rightarrow {m^2} - 5m + 6 = 0$$
$$ \Rightarrow $$ m = 2 or 3
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