119869
If the tangents drawn at the point \(O(0,0)\) and \(P(1+\sqrt{5}, 2)\) on the circle \(x^2+y^2-2 x-4 y=0\) intersect at the point \(Q\), then the area of the triangle \(O P Q\) is equal to:
119870
If \(y=m_1 x+c_1\) and \(y=m_2 x+c_2, m_1 \neq m_2\) are two common tangents of circle \(x^2+y^2=2\) and parabola \(y^2=x\), then the value of \(8\left|m_1 m_2\right|\) is equal to
1 \(3+4 \sqrt{2}\)
2 \(-5+6 \sqrt{2}\)
3 \(-4+3 \sqrt{2}\)
4 \(7+6 \sqrt{2}\)
Explanation:
C \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2=2\) \(\mathrm{C}_2: \mathrm{y}^2=\mathrm{x}\) Here, \(\mathrm{a}=\frac{1}{4}\) \(y=m x+\frac{1}{4 m}\) is also the equation of tangent of circle \(x^2\) \(+y^2=2\) \(\left(\frac{1}{4 \mathrm{~m}}\right)^2=2\left(1+\mathrm{m}^2\right)\) \(32 \mathrm{~m}^4+32 \mathrm{~m}^2-1=0\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32}\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{32 \times 36}}{2 \times 32}=\frac{-32+24 \sqrt{2}}{64}\) \(\mathrm{~m}^2=\frac{-4+3 \sqrt{2}}{8}\) \(\therefore \mathrm{m}_1 \text { and } \mathrm{m}_2=\sqrt{\frac{-4+3 \sqrt{2}}{8}},-\sqrt{\frac{-4+3 \sqrt{2}}{8}}\) \(\mathrm{~m}_1 \cdot \mathrm{m}_2=\sqrt{\frac{(-4+3 \sqrt{2})^2}{(8)^2}}\) \(8\left|\mathrm{~m}_1 \cdot \mathrm{m}_2\right|=8\left[\frac{-4+3 \sqrt{2}}{8}\right]=-4+3 \sqrt{2}\)
JEE Main-25.06.2022
Conic Section
119871
Let a circle \(C\) touch the lines \(L_1: 4 x-3 y+K_1=\) 0 and \(L_2: 4 x-3 y+K_2=0, K_1, K_2 \in R\). If a line passing through the centre of the circle \(C\) intersects \(L_1\) at \((-1,2)\) and \(L_2\) at \((3,-6)\), then the equation of the circle \(C\) is
1 \((x-1)^2+(y-2)^2=4\)
2 \((x+1)^2+(y-2)^2=4\)
3 \((x-1)^2+(y+2)^2=16\)
4 \((x-1)^2+(y-2)^2=16\)
Explanation:
C : \(\mathrm{L}_1: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_1=0\) \(\mathrm{~L}_2: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_2=0\) Now, \(-4-6+\mathrm{K}_1=0 \Rightarrow \mathrm{K}_1=10\) \(12+18+\mathrm{K}_2=0 \Rightarrow \mathrm{K}_2=-30\) \(\Rightarrow\) Tangent to the circle are \(4 \mathrm{x}-3 \mathrm{y}+10=0\) \(4 \mathrm{x}-3 \mathrm{y}-30=0\) Length of diameter \(2 r=\frac{|10+30|}{5}=8\) \(\Rightarrow \mathrm{r}=4\) Now centre is midpoint of \(A \& B\) \(\mathrm{x}=1, \mathrm{y}=-2\) Equation of circle, \((x-1)^2+(y+2)^2=16\)
JEE Main-25.06.2022
Conic Section
119872
The equations of the tangent to the circle \(5 x^2+\) \(5 y^2=1\) parallel to the line \(3 x+4 y=1\) are
1 \(3 x+4 y= \pm 2 \sqrt{5}\)
2 \(3 x+4 y= \pm \sqrt{5}\)
3 \(6 x+8 y= \pm \sqrt{5}\)
4 \(3 x+4 y= \pm 3 \sqrt{5}\)
Explanation:
B Circle, \(5 x^2+5 y^2=1\) \(x^2+y^2=\left(\frac{1}{\sqrt{5}}\right)^2\) \(\therefore\) centre \((0,0), r=\frac{1}{\sqrt{5}}\) Given, line is \(3 x+4 y=1\) \(\mathrm{m}=-\frac{3}{4}\) \(\mathrm{Eq}^{\mathrm{n}}\). of tangent is, \(y=m x \pm a \sqrt{1+m^2}\) Since, required line is parallel to given line, hence it will have same slope and radius:- \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}\) \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}\) \(y=-\frac{-3 x}{4} \pm \frac{1}{\sqrt{5}} \times \frac{5}{4}\) \(4 y=-3 x \pm \sqrt{5}\) \(3 x+4 y= \pm \sqrt{5}\)
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Conic Section
119869
If the tangents drawn at the point \(O(0,0)\) and \(P(1+\sqrt{5}, 2)\) on the circle \(x^2+y^2-2 x-4 y=0\) intersect at the point \(Q\), then the area of the triangle \(O P Q\) is equal to:
119870
If \(y=m_1 x+c_1\) and \(y=m_2 x+c_2, m_1 \neq m_2\) are two common tangents of circle \(x^2+y^2=2\) and parabola \(y^2=x\), then the value of \(8\left|m_1 m_2\right|\) is equal to
1 \(3+4 \sqrt{2}\)
2 \(-5+6 \sqrt{2}\)
3 \(-4+3 \sqrt{2}\)
4 \(7+6 \sqrt{2}\)
Explanation:
C \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2=2\) \(\mathrm{C}_2: \mathrm{y}^2=\mathrm{x}\) Here, \(\mathrm{a}=\frac{1}{4}\) \(y=m x+\frac{1}{4 m}\) is also the equation of tangent of circle \(x^2\) \(+y^2=2\) \(\left(\frac{1}{4 \mathrm{~m}}\right)^2=2\left(1+\mathrm{m}^2\right)\) \(32 \mathrm{~m}^4+32 \mathrm{~m}^2-1=0\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32}\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{32 \times 36}}{2 \times 32}=\frac{-32+24 \sqrt{2}}{64}\) \(\mathrm{~m}^2=\frac{-4+3 \sqrt{2}}{8}\) \(\therefore \mathrm{m}_1 \text { and } \mathrm{m}_2=\sqrt{\frac{-4+3 \sqrt{2}}{8}},-\sqrt{\frac{-4+3 \sqrt{2}}{8}}\) \(\mathrm{~m}_1 \cdot \mathrm{m}_2=\sqrt{\frac{(-4+3 \sqrt{2})^2}{(8)^2}}\) \(8\left|\mathrm{~m}_1 \cdot \mathrm{m}_2\right|=8\left[\frac{-4+3 \sqrt{2}}{8}\right]=-4+3 \sqrt{2}\)
JEE Main-25.06.2022
Conic Section
119871
Let a circle \(C\) touch the lines \(L_1: 4 x-3 y+K_1=\) 0 and \(L_2: 4 x-3 y+K_2=0, K_1, K_2 \in R\). If a line passing through the centre of the circle \(C\) intersects \(L_1\) at \((-1,2)\) and \(L_2\) at \((3,-6)\), then the equation of the circle \(C\) is
1 \((x-1)^2+(y-2)^2=4\)
2 \((x+1)^2+(y-2)^2=4\)
3 \((x-1)^2+(y+2)^2=16\)
4 \((x-1)^2+(y-2)^2=16\)
Explanation:
C : \(\mathrm{L}_1: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_1=0\) \(\mathrm{~L}_2: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_2=0\) Now, \(-4-6+\mathrm{K}_1=0 \Rightarrow \mathrm{K}_1=10\) \(12+18+\mathrm{K}_2=0 \Rightarrow \mathrm{K}_2=-30\) \(\Rightarrow\) Tangent to the circle are \(4 \mathrm{x}-3 \mathrm{y}+10=0\) \(4 \mathrm{x}-3 \mathrm{y}-30=0\) Length of diameter \(2 r=\frac{|10+30|}{5}=8\) \(\Rightarrow \mathrm{r}=4\) Now centre is midpoint of \(A \& B\) \(\mathrm{x}=1, \mathrm{y}=-2\) Equation of circle, \((x-1)^2+(y+2)^2=16\)
JEE Main-25.06.2022
Conic Section
119872
The equations of the tangent to the circle \(5 x^2+\) \(5 y^2=1\) parallel to the line \(3 x+4 y=1\) are
1 \(3 x+4 y= \pm 2 \sqrt{5}\)
2 \(3 x+4 y= \pm \sqrt{5}\)
3 \(6 x+8 y= \pm \sqrt{5}\)
4 \(3 x+4 y= \pm 3 \sqrt{5}\)
Explanation:
B Circle, \(5 x^2+5 y^2=1\) \(x^2+y^2=\left(\frac{1}{\sqrt{5}}\right)^2\) \(\therefore\) centre \((0,0), r=\frac{1}{\sqrt{5}}\) Given, line is \(3 x+4 y=1\) \(\mathrm{m}=-\frac{3}{4}\) \(\mathrm{Eq}^{\mathrm{n}}\). of tangent is, \(y=m x \pm a \sqrt{1+m^2}\) Since, required line is parallel to given line, hence it will have same slope and radius:- \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}\) \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}\) \(y=-\frac{-3 x}{4} \pm \frac{1}{\sqrt{5}} \times \frac{5}{4}\) \(4 y=-3 x \pm \sqrt{5}\) \(3 x+4 y= \pm \sqrt{5}\)
119869
If the tangents drawn at the point \(O(0,0)\) and \(P(1+\sqrt{5}, 2)\) on the circle \(x^2+y^2-2 x-4 y=0\) intersect at the point \(Q\), then the area of the triangle \(O P Q\) is equal to:
119870
If \(y=m_1 x+c_1\) and \(y=m_2 x+c_2, m_1 \neq m_2\) are two common tangents of circle \(x^2+y^2=2\) and parabola \(y^2=x\), then the value of \(8\left|m_1 m_2\right|\) is equal to
1 \(3+4 \sqrt{2}\)
2 \(-5+6 \sqrt{2}\)
3 \(-4+3 \sqrt{2}\)
4 \(7+6 \sqrt{2}\)
Explanation:
C \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2=2\) \(\mathrm{C}_2: \mathrm{y}^2=\mathrm{x}\) Here, \(\mathrm{a}=\frac{1}{4}\) \(y=m x+\frac{1}{4 m}\) is also the equation of tangent of circle \(x^2\) \(+y^2=2\) \(\left(\frac{1}{4 \mathrm{~m}}\right)^2=2\left(1+\mathrm{m}^2\right)\) \(32 \mathrm{~m}^4+32 \mathrm{~m}^2-1=0\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32}\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{32 \times 36}}{2 \times 32}=\frac{-32+24 \sqrt{2}}{64}\) \(\mathrm{~m}^2=\frac{-4+3 \sqrt{2}}{8}\) \(\therefore \mathrm{m}_1 \text { and } \mathrm{m}_2=\sqrt{\frac{-4+3 \sqrt{2}}{8}},-\sqrt{\frac{-4+3 \sqrt{2}}{8}}\) \(\mathrm{~m}_1 \cdot \mathrm{m}_2=\sqrt{\frac{(-4+3 \sqrt{2})^2}{(8)^2}}\) \(8\left|\mathrm{~m}_1 \cdot \mathrm{m}_2\right|=8\left[\frac{-4+3 \sqrt{2}}{8}\right]=-4+3 \sqrt{2}\)
JEE Main-25.06.2022
Conic Section
119871
Let a circle \(C\) touch the lines \(L_1: 4 x-3 y+K_1=\) 0 and \(L_2: 4 x-3 y+K_2=0, K_1, K_2 \in R\). If a line passing through the centre of the circle \(C\) intersects \(L_1\) at \((-1,2)\) and \(L_2\) at \((3,-6)\), then the equation of the circle \(C\) is
1 \((x-1)^2+(y-2)^2=4\)
2 \((x+1)^2+(y-2)^2=4\)
3 \((x-1)^2+(y+2)^2=16\)
4 \((x-1)^2+(y-2)^2=16\)
Explanation:
C : \(\mathrm{L}_1: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_1=0\) \(\mathrm{~L}_2: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_2=0\) Now, \(-4-6+\mathrm{K}_1=0 \Rightarrow \mathrm{K}_1=10\) \(12+18+\mathrm{K}_2=0 \Rightarrow \mathrm{K}_2=-30\) \(\Rightarrow\) Tangent to the circle are \(4 \mathrm{x}-3 \mathrm{y}+10=0\) \(4 \mathrm{x}-3 \mathrm{y}-30=0\) Length of diameter \(2 r=\frac{|10+30|}{5}=8\) \(\Rightarrow \mathrm{r}=4\) Now centre is midpoint of \(A \& B\) \(\mathrm{x}=1, \mathrm{y}=-2\) Equation of circle, \((x-1)^2+(y+2)^2=16\)
JEE Main-25.06.2022
Conic Section
119872
The equations of the tangent to the circle \(5 x^2+\) \(5 y^2=1\) parallel to the line \(3 x+4 y=1\) are
1 \(3 x+4 y= \pm 2 \sqrt{5}\)
2 \(3 x+4 y= \pm \sqrt{5}\)
3 \(6 x+8 y= \pm \sqrt{5}\)
4 \(3 x+4 y= \pm 3 \sqrt{5}\)
Explanation:
B Circle, \(5 x^2+5 y^2=1\) \(x^2+y^2=\left(\frac{1}{\sqrt{5}}\right)^2\) \(\therefore\) centre \((0,0), r=\frac{1}{\sqrt{5}}\) Given, line is \(3 x+4 y=1\) \(\mathrm{m}=-\frac{3}{4}\) \(\mathrm{Eq}^{\mathrm{n}}\). of tangent is, \(y=m x \pm a \sqrt{1+m^2}\) Since, required line is parallel to given line, hence it will have same slope and radius:- \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}\) \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}\) \(y=-\frac{-3 x}{4} \pm \frac{1}{\sqrt{5}} \times \frac{5}{4}\) \(4 y=-3 x \pm \sqrt{5}\) \(3 x+4 y= \pm \sqrt{5}\)
119869
If the tangents drawn at the point \(O(0,0)\) and \(P(1+\sqrt{5}, 2)\) on the circle \(x^2+y^2-2 x-4 y=0\) intersect at the point \(Q\), then the area of the triangle \(O P Q\) is equal to:
119870
If \(y=m_1 x+c_1\) and \(y=m_2 x+c_2, m_1 \neq m_2\) are two common tangents of circle \(x^2+y^2=2\) and parabola \(y^2=x\), then the value of \(8\left|m_1 m_2\right|\) is equal to
1 \(3+4 \sqrt{2}\)
2 \(-5+6 \sqrt{2}\)
3 \(-4+3 \sqrt{2}\)
4 \(7+6 \sqrt{2}\)
Explanation:
C \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2=2\) \(\mathrm{C}_2: \mathrm{y}^2=\mathrm{x}\) Here, \(\mathrm{a}=\frac{1}{4}\) \(y=m x+\frac{1}{4 m}\) is also the equation of tangent of circle \(x^2\) \(+y^2=2\) \(\left(\frac{1}{4 \mathrm{~m}}\right)^2=2\left(1+\mathrm{m}^2\right)\) \(32 \mathrm{~m}^4+32 \mathrm{~m}^2-1=0\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{(32)^2-4(32)(-1)}}{2 \times 32}\) \(\mathrm{~m}^2=\frac{-32 \pm \sqrt{32 \times 36}}{2 \times 32}=\frac{-32+24 \sqrt{2}}{64}\) \(\mathrm{~m}^2=\frac{-4+3 \sqrt{2}}{8}\) \(\therefore \mathrm{m}_1 \text { and } \mathrm{m}_2=\sqrt{\frac{-4+3 \sqrt{2}}{8}},-\sqrt{\frac{-4+3 \sqrt{2}}{8}}\) \(\mathrm{~m}_1 \cdot \mathrm{m}_2=\sqrt{\frac{(-4+3 \sqrt{2})^2}{(8)^2}}\) \(8\left|\mathrm{~m}_1 \cdot \mathrm{m}_2\right|=8\left[\frac{-4+3 \sqrt{2}}{8}\right]=-4+3 \sqrt{2}\)
JEE Main-25.06.2022
Conic Section
119871
Let a circle \(C\) touch the lines \(L_1: 4 x-3 y+K_1=\) 0 and \(L_2: 4 x-3 y+K_2=0, K_1, K_2 \in R\). If a line passing through the centre of the circle \(C\) intersects \(L_1\) at \((-1,2)\) and \(L_2\) at \((3,-6)\), then the equation of the circle \(C\) is
1 \((x-1)^2+(y-2)^2=4\)
2 \((x+1)^2+(y-2)^2=4\)
3 \((x-1)^2+(y+2)^2=16\)
4 \((x-1)^2+(y-2)^2=16\)
Explanation:
C : \(\mathrm{L}_1: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_1=0\) \(\mathrm{~L}_2: 4 \mathrm{x}-3 \mathrm{y}+\mathrm{K}_2=0\) Now, \(-4-6+\mathrm{K}_1=0 \Rightarrow \mathrm{K}_1=10\) \(12+18+\mathrm{K}_2=0 \Rightarrow \mathrm{K}_2=-30\) \(\Rightarrow\) Tangent to the circle are \(4 \mathrm{x}-3 \mathrm{y}+10=0\) \(4 \mathrm{x}-3 \mathrm{y}-30=0\) Length of diameter \(2 r=\frac{|10+30|}{5}=8\) \(\Rightarrow \mathrm{r}=4\) Now centre is midpoint of \(A \& B\) \(\mathrm{x}=1, \mathrm{y}=-2\) Equation of circle, \((x-1)^2+(y+2)^2=16\)
JEE Main-25.06.2022
Conic Section
119872
The equations of the tangent to the circle \(5 x^2+\) \(5 y^2=1\) parallel to the line \(3 x+4 y=1\) are
1 \(3 x+4 y= \pm 2 \sqrt{5}\)
2 \(3 x+4 y= \pm \sqrt{5}\)
3 \(6 x+8 y= \pm \sqrt{5}\)
4 \(3 x+4 y= \pm 3 \sqrt{5}\)
Explanation:
B Circle, \(5 x^2+5 y^2=1\) \(x^2+y^2=\left(\frac{1}{\sqrt{5}}\right)^2\) \(\therefore\) centre \((0,0), r=\frac{1}{\sqrt{5}}\) Given, line is \(3 x+4 y=1\) \(\mathrm{m}=-\frac{3}{4}\) \(\mathrm{Eq}^{\mathrm{n}}\). of tangent is, \(y=m x \pm a \sqrt{1+m^2}\) Since, required line is parallel to given line, hence it will have same slope and radius:- \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}\) \(y=-\frac{3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}\) \(y=-\frac{-3 x}{4} \pm \frac{1}{\sqrt{5}} \times \frac{5}{4}\) \(4 y=-3 x \pm \sqrt{5}\) \(3 x+4 y= \pm \sqrt{5}\)
