# CAT 2023 Question Paper | Quant Slot 1

###### CAT Previous Year Paper | CAT Quant Questions | Question 17

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 17 : Let $$mathrm{C}$ be the circle $x^2+y^2+4 x-6 y-3=0$ and $$mathrm{L}$ be the locus of the point of intersection of a pair of tangents to $\mathrm{C}$ with the angle between the two tangents equal to $60^{\circ}$. Then, the point at which $L$ touches the line $x=6$ is 1.$6,4$
2. (6,8)
3. (6,3)
4. (6,6)

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$x ^ { 2 } + y ^ { 2 } + 4 x - 6 y - 3 = 0$
$x ^ { 2 } + 2 $cdot 2 $cdot x + 2 ^ { 2 } - 2 ^ { 2 } + y ^ { 2 } - 2 \cdot 3 \cdot y + 3 ^ { 2 } - 3 ^ { 2 } - 3 = 0$ $$ x + 2$ ^ { 2 } + ( y - 3 ) ^ { 2 } - $left$ 2 ^ { 2 } + 3 ^ { 2 } + 3 $right) = 0$ $$ x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 4 ^ { 2 }$
This is the equation of a circle with radius 4 units and centered at (-2, 3)
From a point L we drop two tangents on the circle such that the angle between the tangents is 60o.

$$angle L P O = 90 ^ { $circ }$ $\angle P L O = 30 ^ { \circ }$ $\sin \left$ 30 ^ { $circ } \right$ = $frac { 1 } { 2 } = \frac { P O } { O L } = \frac { 4 } { 4 + x }$ $4 + x = 2$ 4$$
$x = 4$
Therefore the locus of the point L, is a circle centered at (-2, 3) and has a radius of (4 + x = 8) units.
The equation of this locus is thus, $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 8 ^ { 2 }$
When x = 6, we have, $( 8 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 8 ^ { 2 }$, that is y = 3
The circle , $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 8 ^ { 2 }$, touches the line $x = 6$ at (6, 3).

##### The answer is '(6,3)'

Choice C is the correct answer.

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