119904
The circle \(x^2+y^2=4 x+8 y+5\) intersects the line \(3 x-4 y=m\) at two distinct points, if
1 \(-85\lt \mathrm{m}\lt -35\)
2 \(-35\lt \mathrm{m}\lt 15\)
3 \(15\lt \mathrm{m}\lt 65\)
4 \(35\lt \mathrm{m}\lt 85\) [-2010]
Explanation:
B Given circle equation \(x^2+y^2=4 x+8 y+5\) \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=(5)^2\) \(\text { Therefore, }\) centre of circle \(=(2,4)\) Radius of circle \(=5\) Given that the circle is intersecting the line \(3 x-4 y=m\) at two distinct points Therefore length of perpendicular \(\lt \) Radius of circle \(\frac{|6-16-\mathrm{m}|}{\sqrt{(3)^2+(4)^2}}\lt 5\) \(\frac{|-10-\mathrm{m}|}{\sqrt{9+16}}\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)
Conic Section
119905
If a circle \(C\) passing through the point \((4,0)\) touches the circle \(x^2+y^2+4 x-6 y=12\) externally at the point \((1,-1)\), then radius of \(C\) is
1 5
2 \(2 \sqrt{5}\)
3 \(\sqrt{57}\)
4 4
Explanation:
A Given circle equation \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}=12\) Equation of tangent at \((1,-1)\) \(\mathrm{x}(1)+\mathrm{y}(-1)+2(\mathrm{x}+1)-3(\mathrm{y}-1)-12=0\) \(\mathrm{x}-\mathrm{y}+2 \mathrm{x}+2-3 \mathrm{y}+3-12=0\) \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(3 \mathrm{x}-4 \mathrm{y}=7\) \(\therefore\) equation of circle is \((x-1)^2+(y+1)^2+\lambda(3 x-4 y-7)=0\) Which passes through \((4,0)\) \(9+1+\lambda(12-7)=0\) \(\lambda=-2\) Put the value of \(\lambda\) in equation (i) we get, \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}+10 \mathrm{y}+16=0\) \(\text { Radius }=\sqrt{16+25-16}=\sqrt{25}\) \(\mathrm{R}=5\)
JEE Main 10.01.2019
Conic Section
119906
The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(B\). A circle is drawn through A, B and the origin. Then, the sum of perpendicular distances from \(A\) and \(B\) on the tangent to the circle at the origin is
1 \(2 \sqrt{5}\)
2 \(\frac{\sqrt{5}}{4}\)
3 \(4 \sqrt{5}\)
4 \(\frac{\sqrt{5}}{2}\)
Explanation:
D Equation of circle- \((\mathrm{x}-)(\mathrm{x}-0)+(\mathrm{y}-0)\left(\mathrm{y}-\frac{1}{2}\right)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\frac{\mathrm{y}}{2}=0\) \(\text { Equation of tangent at origin is } 2 \mathrm{x}+\mathrm{y}=0\) \(\text { Distance between point 'A' and tangent - }\) \(\mathrm{AM}=\frac{|2 \times 1+1 \times 0|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\) Distance between point ' \(\mathrm{B}\) ' and tangent - \(\mathrm{BN}=\frac{\left|2 \times 0+1 \times \frac{1}{2}\right|}{\sqrt{5}}=\frac{1}{2 \sqrt{5}}\) From equation the sum of perpendicular distance, \(\mathrm{AM}+\mathrm{BN}=\frac{2}{\sqrt{5}}+\frac{1}{2 \sqrt{5}}=\frac{4+1}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}\)
JEE Main 11.01.2019
Conic Section
119907
Equation of a common tangent to the circle, \(x^2\) \(+y^2-6 x=0\) and the parabola, \(y^2=4 x\), is
1 \(\sqrt{3} y=3 x+1\)
2 \(2 \sqrt{3} y=12 x+1\)
3 \(\sqrt{3} y=x+3\)
4 \(2 \sqrt{3} y=-x-12\)
Explanation:
C Given, equation of tangent to the parabola \(y^2=4 x \text { is }\) \(y=m x+\frac{1}{m}\) \(\text { Equation of tangent to the circle - }\) \(y=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(\text { Where, } m \text { is the slope }\) \(\text { Since, two curve have common tangents }\) \(\therefore m x+\frac{1}{m}=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(m= \pm \frac{1}{\sqrt{3}}\) \(\text { Tangents are } x+\sqrt{3} y+3=0\) \(\text { and } x-\sqrt{3} y+3=0\) \(\text { or } \sqrt{3} y=x+3\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119904
The circle \(x^2+y^2=4 x+8 y+5\) intersects the line \(3 x-4 y=m\) at two distinct points, if
1 \(-85\lt \mathrm{m}\lt -35\)
2 \(-35\lt \mathrm{m}\lt 15\)
3 \(15\lt \mathrm{m}\lt 65\)
4 \(35\lt \mathrm{m}\lt 85\) [-2010]
Explanation:
B Given circle equation \(x^2+y^2=4 x+8 y+5\) \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=(5)^2\) \(\text { Therefore, }\) centre of circle \(=(2,4)\) Radius of circle \(=5\) Given that the circle is intersecting the line \(3 x-4 y=m\) at two distinct points Therefore length of perpendicular \(\lt \) Radius of circle \(\frac{|6-16-\mathrm{m}|}{\sqrt{(3)^2+(4)^2}}\lt 5\) \(\frac{|-10-\mathrm{m}|}{\sqrt{9+16}}\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)
Conic Section
119905
If a circle \(C\) passing through the point \((4,0)\) touches the circle \(x^2+y^2+4 x-6 y=12\) externally at the point \((1,-1)\), then radius of \(C\) is
1 5
2 \(2 \sqrt{5}\)
3 \(\sqrt{57}\)
4 4
Explanation:
A Given circle equation \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}=12\) Equation of tangent at \((1,-1)\) \(\mathrm{x}(1)+\mathrm{y}(-1)+2(\mathrm{x}+1)-3(\mathrm{y}-1)-12=0\) \(\mathrm{x}-\mathrm{y}+2 \mathrm{x}+2-3 \mathrm{y}+3-12=0\) \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(3 \mathrm{x}-4 \mathrm{y}=7\) \(\therefore\) equation of circle is \((x-1)^2+(y+1)^2+\lambda(3 x-4 y-7)=0\) Which passes through \((4,0)\) \(9+1+\lambda(12-7)=0\) \(\lambda=-2\) Put the value of \(\lambda\) in equation (i) we get, \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}+10 \mathrm{y}+16=0\) \(\text { Radius }=\sqrt{16+25-16}=\sqrt{25}\) \(\mathrm{R}=5\)
JEE Main 10.01.2019
Conic Section
119906
The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(B\). A circle is drawn through A, B and the origin. Then, the sum of perpendicular distances from \(A\) and \(B\) on the tangent to the circle at the origin is
1 \(2 \sqrt{5}\)
2 \(\frac{\sqrt{5}}{4}\)
3 \(4 \sqrt{5}\)
4 \(\frac{\sqrt{5}}{2}\)
Explanation:
D Equation of circle- \((\mathrm{x}-)(\mathrm{x}-0)+(\mathrm{y}-0)\left(\mathrm{y}-\frac{1}{2}\right)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\frac{\mathrm{y}}{2}=0\) \(\text { Equation of tangent at origin is } 2 \mathrm{x}+\mathrm{y}=0\) \(\text { Distance between point 'A' and tangent - }\) \(\mathrm{AM}=\frac{|2 \times 1+1 \times 0|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\) Distance between point ' \(\mathrm{B}\) ' and tangent - \(\mathrm{BN}=\frac{\left|2 \times 0+1 \times \frac{1}{2}\right|}{\sqrt{5}}=\frac{1}{2 \sqrt{5}}\) From equation the sum of perpendicular distance, \(\mathrm{AM}+\mathrm{BN}=\frac{2}{\sqrt{5}}+\frac{1}{2 \sqrt{5}}=\frac{4+1}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}\)
JEE Main 11.01.2019
Conic Section
119907
Equation of a common tangent to the circle, \(x^2\) \(+y^2-6 x=0\) and the parabola, \(y^2=4 x\), is
1 \(\sqrt{3} y=3 x+1\)
2 \(2 \sqrt{3} y=12 x+1\)
3 \(\sqrt{3} y=x+3\)
4 \(2 \sqrt{3} y=-x-12\)
Explanation:
C Given, equation of tangent to the parabola \(y^2=4 x \text { is }\) \(y=m x+\frac{1}{m}\) \(\text { Equation of tangent to the circle - }\) \(y=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(\text { Where, } m \text { is the slope }\) \(\text { Since, two curve have common tangents }\) \(\therefore m x+\frac{1}{m}=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(m= \pm \frac{1}{\sqrt{3}}\) \(\text { Tangents are } x+\sqrt{3} y+3=0\) \(\text { and } x-\sqrt{3} y+3=0\) \(\text { or } \sqrt{3} y=x+3\)
119904
The circle \(x^2+y^2=4 x+8 y+5\) intersects the line \(3 x-4 y=m\) at two distinct points, if
1 \(-85\lt \mathrm{m}\lt -35\)
2 \(-35\lt \mathrm{m}\lt 15\)
3 \(15\lt \mathrm{m}\lt 65\)
4 \(35\lt \mathrm{m}\lt 85\) [-2010]
Explanation:
B Given circle equation \(x^2+y^2=4 x+8 y+5\) \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=(5)^2\) \(\text { Therefore, }\) centre of circle \(=(2,4)\) Radius of circle \(=5\) Given that the circle is intersecting the line \(3 x-4 y=m\) at two distinct points Therefore length of perpendicular \(\lt \) Radius of circle \(\frac{|6-16-\mathrm{m}|}{\sqrt{(3)^2+(4)^2}}\lt 5\) \(\frac{|-10-\mathrm{m}|}{\sqrt{9+16}}\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)
Conic Section
119905
If a circle \(C\) passing through the point \((4,0)\) touches the circle \(x^2+y^2+4 x-6 y=12\) externally at the point \((1,-1)\), then radius of \(C\) is
1 5
2 \(2 \sqrt{5}\)
3 \(\sqrt{57}\)
4 4
Explanation:
A Given circle equation \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}=12\) Equation of tangent at \((1,-1)\) \(\mathrm{x}(1)+\mathrm{y}(-1)+2(\mathrm{x}+1)-3(\mathrm{y}-1)-12=0\) \(\mathrm{x}-\mathrm{y}+2 \mathrm{x}+2-3 \mathrm{y}+3-12=0\) \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(3 \mathrm{x}-4 \mathrm{y}=7\) \(\therefore\) equation of circle is \((x-1)^2+(y+1)^2+\lambda(3 x-4 y-7)=0\) Which passes through \((4,0)\) \(9+1+\lambda(12-7)=0\) \(\lambda=-2\) Put the value of \(\lambda\) in equation (i) we get, \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}+10 \mathrm{y}+16=0\) \(\text { Radius }=\sqrt{16+25-16}=\sqrt{25}\) \(\mathrm{R}=5\)
JEE Main 10.01.2019
Conic Section
119906
The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(B\). A circle is drawn through A, B and the origin. Then, the sum of perpendicular distances from \(A\) and \(B\) on the tangent to the circle at the origin is
1 \(2 \sqrt{5}\)
2 \(\frac{\sqrt{5}}{4}\)
3 \(4 \sqrt{5}\)
4 \(\frac{\sqrt{5}}{2}\)
Explanation:
D Equation of circle- \((\mathrm{x}-)(\mathrm{x}-0)+(\mathrm{y}-0)\left(\mathrm{y}-\frac{1}{2}\right)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\frac{\mathrm{y}}{2}=0\) \(\text { Equation of tangent at origin is } 2 \mathrm{x}+\mathrm{y}=0\) \(\text { Distance between point 'A' and tangent - }\) \(\mathrm{AM}=\frac{|2 \times 1+1 \times 0|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\) Distance between point ' \(\mathrm{B}\) ' and tangent - \(\mathrm{BN}=\frac{\left|2 \times 0+1 \times \frac{1}{2}\right|}{\sqrt{5}}=\frac{1}{2 \sqrt{5}}\) From equation the sum of perpendicular distance, \(\mathrm{AM}+\mathrm{BN}=\frac{2}{\sqrt{5}}+\frac{1}{2 \sqrt{5}}=\frac{4+1}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}\)
JEE Main 11.01.2019
Conic Section
119907
Equation of a common tangent to the circle, \(x^2\) \(+y^2-6 x=0\) and the parabola, \(y^2=4 x\), is
1 \(\sqrt{3} y=3 x+1\)
2 \(2 \sqrt{3} y=12 x+1\)
3 \(\sqrt{3} y=x+3\)
4 \(2 \sqrt{3} y=-x-12\)
Explanation:
C Given, equation of tangent to the parabola \(y^2=4 x \text { is }\) \(y=m x+\frac{1}{m}\) \(\text { Equation of tangent to the circle - }\) \(y=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(\text { Where, } m \text { is the slope }\) \(\text { Since, two curve have common tangents }\) \(\therefore m x+\frac{1}{m}=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(m= \pm \frac{1}{\sqrt{3}}\) \(\text { Tangents are } x+\sqrt{3} y+3=0\) \(\text { and } x-\sqrt{3} y+3=0\) \(\text { or } \sqrt{3} y=x+3\)
119904
The circle \(x^2+y^2=4 x+8 y+5\) intersects the line \(3 x-4 y=m\) at two distinct points, if
1 \(-85\lt \mathrm{m}\lt -35\)
2 \(-35\lt \mathrm{m}\lt 15\)
3 \(15\lt \mathrm{m}\lt 65\)
4 \(35\lt \mathrm{m}\lt 85\) [-2010]
Explanation:
B Given circle equation \(x^2+y^2=4 x+8 y+5\) \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=(5)^2\) \(\text { Therefore, }\) centre of circle \(=(2,4)\) Radius of circle \(=5\) Given that the circle is intersecting the line \(3 x-4 y=m\) at two distinct points Therefore length of perpendicular \(\lt \) Radius of circle \(\frac{|6-16-\mathrm{m}|}{\sqrt{(3)^2+(4)^2}}\lt 5\) \(\frac{|-10-\mathrm{m}|}{\sqrt{9+16}}\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)
Conic Section
119905
If a circle \(C\) passing through the point \((4,0)\) touches the circle \(x^2+y^2+4 x-6 y=12\) externally at the point \((1,-1)\), then radius of \(C\) is
1 5
2 \(2 \sqrt{5}\)
3 \(\sqrt{57}\)
4 4
Explanation:
A Given circle equation \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}=12\) Equation of tangent at \((1,-1)\) \(\mathrm{x}(1)+\mathrm{y}(-1)+2(\mathrm{x}+1)-3(\mathrm{y}-1)-12=0\) \(\mathrm{x}-\mathrm{y}+2 \mathrm{x}+2-3 \mathrm{y}+3-12=0\) \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(3 \mathrm{x}-4 \mathrm{y}=7\) \(\therefore\) equation of circle is \((x-1)^2+(y+1)^2+\lambda(3 x-4 y-7)=0\) Which passes through \((4,0)\) \(9+1+\lambda(12-7)=0\) \(\lambda=-2\) Put the value of \(\lambda\) in equation (i) we get, \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}+10 \mathrm{y}+16=0\) \(\text { Radius }=\sqrt{16+25-16}=\sqrt{25}\) \(\mathrm{R}=5\)
JEE Main 10.01.2019
Conic Section
119906
The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(B\). A circle is drawn through A, B and the origin. Then, the sum of perpendicular distances from \(A\) and \(B\) on the tangent to the circle at the origin is
1 \(2 \sqrt{5}\)
2 \(\frac{\sqrt{5}}{4}\)
3 \(4 \sqrt{5}\)
4 \(\frac{\sqrt{5}}{2}\)
Explanation:
D Equation of circle- \((\mathrm{x}-)(\mathrm{x}-0)+(\mathrm{y}-0)\left(\mathrm{y}-\frac{1}{2}\right)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\frac{\mathrm{y}}{2}=0\) \(\text { Equation of tangent at origin is } 2 \mathrm{x}+\mathrm{y}=0\) \(\text { Distance between point 'A' and tangent - }\) \(\mathrm{AM}=\frac{|2 \times 1+1 \times 0|}{\sqrt{5}}=\frac{2}{\sqrt{5}}\) Distance between point ' \(\mathrm{B}\) ' and tangent - \(\mathrm{BN}=\frac{\left|2 \times 0+1 \times \frac{1}{2}\right|}{\sqrt{5}}=\frac{1}{2 \sqrt{5}}\) From equation the sum of perpendicular distance, \(\mathrm{AM}+\mathrm{BN}=\frac{2}{\sqrt{5}}+\frac{1}{2 \sqrt{5}}=\frac{4+1}{2 \sqrt{5}}=\frac{\sqrt{5}}{2}\)
JEE Main 11.01.2019
Conic Section
119907
Equation of a common tangent to the circle, \(x^2\) \(+y^2-6 x=0\) and the parabola, \(y^2=4 x\), is
1 \(\sqrt{3} y=3 x+1\)
2 \(2 \sqrt{3} y=12 x+1\)
3 \(\sqrt{3} y=x+3\)
4 \(2 \sqrt{3} y=-x-12\)
Explanation:
C Given, equation of tangent to the parabola \(y^2=4 x \text { is }\) \(y=m x+\frac{1}{m}\) \(\text { Equation of tangent to the circle - }\) \(y=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(\text { Where, } m \text { is the slope }\) \(\text { Since, two curve have common tangents }\) \(\therefore m x+\frac{1}{m}=m(x-3)+3 \sqrt{1+\mathrm{m}^2}\) \(m= \pm \frac{1}{\sqrt{3}}\) \(\text { Tangents are } x+\sqrt{3} y+3=0\) \(\text { and } x-\sqrt{3} y+3=0\) \(\text { or } \sqrt{3} y=x+3\)
