119973
If the straight lines \(x-4 y+7=0\) and \(3 x-12 y\) \(+11=0\) are tangents to a circle. Then, the radius of the circle is
1 \(\frac{10}{3 \sqrt{17}}\)
2 \(\frac{5}{3 \sqrt{7}}\)
3 \(\frac{15}{\sqrt{17}}\)
4 \(\frac{5}{3 \sqrt{17}}\)
5 \(\frac{5}{3 \sqrt{13}}\)
Explanation:
D Given, straight line- \(x-4 y+7=0\) \(3 x-12 y+11=0\) \(x-4 y+\frac{11}{3}=0\) For finding the distance \(\mathrm{b} / \mathrm{w}\) any two parallel lines ax + by \(+\mathrm{C}_1\) and ax + by \(+\mathrm{C}_2\) \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(\frac{\left|7-\frac{11}{3}\right|}{\sqrt{1^2+(4)^2}}=\frac{\left|\frac{10}{3}\right|}{\sqrt{17}}=\frac{10}{3 \sqrt{17}}\)So, the radius will be \(=\frac{5}{3 \sqrt{17}}\)
Kerala CEE-2013
Conic Section
119844
The equation of the tangent to the circle \(x^2+y^2\) \(=25\) at \((4,3)\) is
1 \(4 x-3 y=25\)
2 \(4 x+3 y=25\)
3 \(4 x+3 y=126\)
4 \(4 x+3 y=9\)
Explanation:
B Given, \(x^2+y^2=25\) Point \((4,3)\) The slope of tangent is \(2 x+2 y \frac{d y}{d x}=0\) \(2 x=-2 y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{-x}{y}\) \(\left(\frac{d y}{d x}\right)_{(4,3)}=\frac{-4}{3}\) \(\therefore\) Equation of tangent is \(y-y_1=\left(\frac{d y}{d x}\right)_{(a, b)}\left(x-x_1\right)\) \((y-3)=\frac{-4}{3}(x-4)\) \(3 y-9=-4 x+16\) \(4 x+3 y=25\)
JEEE-2018
Conic Section
119845
A tangent is drawn to the circle \(2 x^2+2 y^2-3 x+4 y=0\) at the point ' \(A\) ' and it meets the line \(x+y=3\) at \(B(2,1)\), then \(A B=\)
1 \(\sqrt{10}\)
2 2
3 \(2 \sqrt{2}\)
4 0
Explanation:
B Equation of circle is, \(2 x^2+2 y^2-3 x+4 y=0\) \(x^2+y^2-\frac{3}{2} x+2 y=0\) Here, \(A B\) is the length of tangent to the circle from \(B\) \(\text { i.e. } A B=\sqrt{(2)^2+(1)^2-\frac{3}{2}(2)+2(1)}\) \(=\sqrt{4+1-3+2}=\sqrt{4}\) \(=2 \text {. }\)
Karnataka CET-2013
Conic Section
119852
The sum of the minimum distance and the maximum distance from the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
1 20
2 12
3 10
4 16
Explanation:
A Centre of the given circle \(=\mathrm{C}(-2,5)\) Radius of the circle \(\mathrm{CN}=\mathrm{CT}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{2^2+5^2+7}=\sqrt{36}=6\) Distance between (4,-3) and ( \(-2,5\) ) \(\mathrm{PC}=\sqrt{6^2+8^2}=\sqrt{100}=10\) We join the external point, \((4,-3)\) to the centre of the circle \((-2,5)\). Then PT is the minimum distance, from external point \(\mathrm{P}\) to the circle and \(\mathrm{PN}\) is the maximum distance. Minimum distance \(=\mathrm{PT}=\mathrm{PC}-\mathrm{CT}=10-6=4\) Maximum distance \(=\mathrm{PN}=\mathrm{PC}+\mathrm{CN}=10+6=16\) So, sum of minimum and maximum distance \(=\) \(16+4=20\)
COMEDK-2018
Conic Section
119846
The equations of the two tangents from \((-5,-4)\) to the circle \(x^2+y^2+4 x+6 y+8=0\) are
119973
If the straight lines \(x-4 y+7=0\) and \(3 x-12 y\) \(+11=0\) are tangents to a circle. Then, the radius of the circle is
1 \(\frac{10}{3 \sqrt{17}}\)
2 \(\frac{5}{3 \sqrt{7}}\)
3 \(\frac{15}{\sqrt{17}}\)
4 \(\frac{5}{3 \sqrt{17}}\)
5 \(\frac{5}{3 \sqrt{13}}\)
Explanation:
D Given, straight line- \(x-4 y+7=0\) \(3 x-12 y+11=0\) \(x-4 y+\frac{11}{3}=0\) For finding the distance \(\mathrm{b} / \mathrm{w}\) any two parallel lines ax + by \(+\mathrm{C}_1\) and ax + by \(+\mathrm{C}_2\) \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(\frac{\left|7-\frac{11}{3}\right|}{\sqrt{1^2+(4)^2}}=\frac{\left|\frac{10}{3}\right|}{\sqrt{17}}=\frac{10}{3 \sqrt{17}}\)So, the radius will be \(=\frac{5}{3 \sqrt{17}}\)
Kerala CEE-2013
Conic Section
119844
The equation of the tangent to the circle \(x^2+y^2\) \(=25\) at \((4,3)\) is
1 \(4 x-3 y=25\)
2 \(4 x+3 y=25\)
3 \(4 x+3 y=126\)
4 \(4 x+3 y=9\)
Explanation:
B Given, \(x^2+y^2=25\) Point \((4,3)\) The slope of tangent is \(2 x+2 y \frac{d y}{d x}=0\) \(2 x=-2 y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{-x}{y}\) \(\left(\frac{d y}{d x}\right)_{(4,3)}=\frac{-4}{3}\) \(\therefore\) Equation of tangent is \(y-y_1=\left(\frac{d y}{d x}\right)_{(a, b)}\left(x-x_1\right)\) \((y-3)=\frac{-4}{3}(x-4)\) \(3 y-9=-4 x+16\) \(4 x+3 y=25\)
JEEE-2018
Conic Section
119845
A tangent is drawn to the circle \(2 x^2+2 y^2-3 x+4 y=0\) at the point ' \(A\) ' and it meets the line \(x+y=3\) at \(B(2,1)\), then \(A B=\)
1 \(\sqrt{10}\)
2 2
3 \(2 \sqrt{2}\)
4 0
Explanation:
B Equation of circle is, \(2 x^2+2 y^2-3 x+4 y=0\) \(x^2+y^2-\frac{3}{2} x+2 y=0\) Here, \(A B\) is the length of tangent to the circle from \(B\) \(\text { i.e. } A B=\sqrt{(2)^2+(1)^2-\frac{3}{2}(2)+2(1)}\) \(=\sqrt{4+1-3+2}=\sqrt{4}\) \(=2 \text {. }\)
Karnataka CET-2013
Conic Section
119852
The sum of the minimum distance and the maximum distance from the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
1 20
2 12
3 10
4 16
Explanation:
A Centre of the given circle \(=\mathrm{C}(-2,5)\) Radius of the circle \(\mathrm{CN}=\mathrm{CT}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{2^2+5^2+7}=\sqrt{36}=6\) Distance between (4,-3) and ( \(-2,5\) ) \(\mathrm{PC}=\sqrt{6^2+8^2}=\sqrt{100}=10\) We join the external point, \((4,-3)\) to the centre of the circle \((-2,5)\). Then PT is the minimum distance, from external point \(\mathrm{P}\) to the circle and \(\mathrm{PN}\) is the maximum distance. Minimum distance \(=\mathrm{PT}=\mathrm{PC}-\mathrm{CT}=10-6=4\) Maximum distance \(=\mathrm{PN}=\mathrm{PC}+\mathrm{CN}=10+6=16\) So, sum of minimum and maximum distance \(=\) \(16+4=20\)
COMEDK-2018
Conic Section
119846
The equations of the two tangents from \((-5,-4)\) to the circle \(x^2+y^2+4 x+6 y+8=0\) are
119973
If the straight lines \(x-4 y+7=0\) and \(3 x-12 y\) \(+11=0\) are tangents to a circle. Then, the radius of the circle is
1 \(\frac{10}{3 \sqrt{17}}\)
2 \(\frac{5}{3 \sqrt{7}}\)
3 \(\frac{15}{\sqrt{17}}\)
4 \(\frac{5}{3 \sqrt{17}}\)
5 \(\frac{5}{3 \sqrt{13}}\)
Explanation:
D Given, straight line- \(x-4 y+7=0\) \(3 x-12 y+11=0\) \(x-4 y+\frac{11}{3}=0\) For finding the distance \(\mathrm{b} / \mathrm{w}\) any two parallel lines ax + by \(+\mathrm{C}_1\) and ax + by \(+\mathrm{C}_2\) \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(\frac{\left|7-\frac{11}{3}\right|}{\sqrt{1^2+(4)^2}}=\frac{\left|\frac{10}{3}\right|}{\sqrt{17}}=\frac{10}{3 \sqrt{17}}\)So, the radius will be \(=\frac{5}{3 \sqrt{17}}\)
Kerala CEE-2013
Conic Section
119844
The equation of the tangent to the circle \(x^2+y^2\) \(=25\) at \((4,3)\) is
1 \(4 x-3 y=25\)
2 \(4 x+3 y=25\)
3 \(4 x+3 y=126\)
4 \(4 x+3 y=9\)
Explanation:
B Given, \(x^2+y^2=25\) Point \((4,3)\) The slope of tangent is \(2 x+2 y \frac{d y}{d x}=0\) \(2 x=-2 y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{-x}{y}\) \(\left(\frac{d y}{d x}\right)_{(4,3)}=\frac{-4}{3}\) \(\therefore\) Equation of tangent is \(y-y_1=\left(\frac{d y}{d x}\right)_{(a, b)}\left(x-x_1\right)\) \((y-3)=\frac{-4}{3}(x-4)\) \(3 y-9=-4 x+16\) \(4 x+3 y=25\)
JEEE-2018
Conic Section
119845
A tangent is drawn to the circle \(2 x^2+2 y^2-3 x+4 y=0\) at the point ' \(A\) ' and it meets the line \(x+y=3\) at \(B(2,1)\), then \(A B=\)
1 \(\sqrt{10}\)
2 2
3 \(2 \sqrt{2}\)
4 0
Explanation:
B Equation of circle is, \(2 x^2+2 y^2-3 x+4 y=0\) \(x^2+y^2-\frac{3}{2} x+2 y=0\) Here, \(A B\) is the length of tangent to the circle from \(B\) \(\text { i.e. } A B=\sqrt{(2)^2+(1)^2-\frac{3}{2}(2)+2(1)}\) \(=\sqrt{4+1-3+2}=\sqrt{4}\) \(=2 \text {. }\)
Karnataka CET-2013
Conic Section
119852
The sum of the minimum distance and the maximum distance from the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
1 20
2 12
3 10
4 16
Explanation:
A Centre of the given circle \(=\mathrm{C}(-2,5)\) Radius of the circle \(\mathrm{CN}=\mathrm{CT}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{2^2+5^2+7}=\sqrt{36}=6\) Distance between (4,-3) and ( \(-2,5\) ) \(\mathrm{PC}=\sqrt{6^2+8^2}=\sqrt{100}=10\) We join the external point, \((4,-3)\) to the centre of the circle \((-2,5)\). Then PT is the minimum distance, from external point \(\mathrm{P}\) to the circle and \(\mathrm{PN}\) is the maximum distance. Minimum distance \(=\mathrm{PT}=\mathrm{PC}-\mathrm{CT}=10-6=4\) Maximum distance \(=\mathrm{PN}=\mathrm{PC}+\mathrm{CN}=10+6=16\) So, sum of minimum and maximum distance \(=\) \(16+4=20\)
COMEDK-2018
Conic Section
119846
The equations of the two tangents from \((-5,-4)\) to the circle \(x^2+y^2+4 x+6 y+8=0\) are
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Conic Section
119973
If the straight lines \(x-4 y+7=0\) and \(3 x-12 y\) \(+11=0\) are tangents to a circle. Then, the radius of the circle is
1 \(\frac{10}{3 \sqrt{17}}\)
2 \(\frac{5}{3 \sqrt{7}}\)
3 \(\frac{15}{\sqrt{17}}\)
4 \(\frac{5}{3 \sqrt{17}}\)
5 \(\frac{5}{3 \sqrt{13}}\)
Explanation:
D Given, straight line- \(x-4 y+7=0\) \(3 x-12 y+11=0\) \(x-4 y+\frac{11}{3}=0\) For finding the distance \(\mathrm{b} / \mathrm{w}\) any two parallel lines ax + by \(+\mathrm{C}_1\) and ax + by \(+\mathrm{C}_2\) \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(\frac{\left|7-\frac{11}{3}\right|}{\sqrt{1^2+(4)^2}}=\frac{\left|\frac{10}{3}\right|}{\sqrt{17}}=\frac{10}{3 \sqrt{17}}\)So, the radius will be \(=\frac{5}{3 \sqrt{17}}\)
Kerala CEE-2013
Conic Section
119844
The equation of the tangent to the circle \(x^2+y^2\) \(=25\) at \((4,3)\) is
1 \(4 x-3 y=25\)
2 \(4 x+3 y=25\)
3 \(4 x+3 y=126\)
4 \(4 x+3 y=9\)
Explanation:
B Given, \(x^2+y^2=25\) Point \((4,3)\) The slope of tangent is \(2 x+2 y \frac{d y}{d x}=0\) \(2 x=-2 y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{-x}{y}\) \(\left(\frac{d y}{d x}\right)_{(4,3)}=\frac{-4}{3}\) \(\therefore\) Equation of tangent is \(y-y_1=\left(\frac{d y}{d x}\right)_{(a, b)}\left(x-x_1\right)\) \((y-3)=\frac{-4}{3}(x-4)\) \(3 y-9=-4 x+16\) \(4 x+3 y=25\)
JEEE-2018
Conic Section
119845
A tangent is drawn to the circle \(2 x^2+2 y^2-3 x+4 y=0\) at the point ' \(A\) ' and it meets the line \(x+y=3\) at \(B(2,1)\), then \(A B=\)
1 \(\sqrt{10}\)
2 2
3 \(2 \sqrt{2}\)
4 0
Explanation:
B Equation of circle is, \(2 x^2+2 y^2-3 x+4 y=0\) \(x^2+y^2-\frac{3}{2} x+2 y=0\) Here, \(A B\) is the length of tangent to the circle from \(B\) \(\text { i.e. } A B=\sqrt{(2)^2+(1)^2-\frac{3}{2}(2)+2(1)}\) \(=\sqrt{4+1-3+2}=\sqrt{4}\) \(=2 \text {. }\)
Karnataka CET-2013
Conic Section
119852
The sum of the minimum distance and the maximum distance from the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
1 20
2 12
3 10
4 16
Explanation:
A Centre of the given circle \(=\mathrm{C}(-2,5)\) Radius of the circle \(\mathrm{CN}=\mathrm{CT}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{2^2+5^2+7}=\sqrt{36}=6\) Distance between (4,-3) and ( \(-2,5\) ) \(\mathrm{PC}=\sqrt{6^2+8^2}=\sqrt{100}=10\) We join the external point, \((4,-3)\) to the centre of the circle \((-2,5)\). Then PT is the minimum distance, from external point \(\mathrm{P}\) to the circle and \(\mathrm{PN}\) is the maximum distance. Minimum distance \(=\mathrm{PT}=\mathrm{PC}-\mathrm{CT}=10-6=4\) Maximum distance \(=\mathrm{PN}=\mathrm{PC}+\mathrm{CN}=10+6=16\) So, sum of minimum and maximum distance \(=\) \(16+4=20\)
COMEDK-2018
Conic Section
119846
The equations of the two tangents from \((-5,-4)\) to the circle \(x^2+y^2+4 x+6 y+8=0\) are
119973
If the straight lines \(x-4 y+7=0\) and \(3 x-12 y\) \(+11=0\) are tangents to a circle. Then, the radius of the circle is
1 \(\frac{10}{3 \sqrt{17}}\)
2 \(\frac{5}{3 \sqrt{7}}\)
3 \(\frac{15}{\sqrt{17}}\)
4 \(\frac{5}{3 \sqrt{17}}\)
5 \(\frac{5}{3 \sqrt{13}}\)
Explanation:
D Given, straight line- \(x-4 y+7=0\) \(3 x-12 y+11=0\) \(x-4 y+\frac{11}{3}=0\) For finding the distance \(\mathrm{b} / \mathrm{w}\) any two parallel lines ax + by \(+\mathrm{C}_1\) and ax + by \(+\mathrm{C}_2\) \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\) \(\frac{\left|7-\frac{11}{3}\right|}{\sqrt{1^2+(4)^2}}=\frac{\left|\frac{10}{3}\right|}{\sqrt{17}}=\frac{10}{3 \sqrt{17}}\)So, the radius will be \(=\frac{5}{3 \sqrt{17}}\)
Kerala CEE-2013
Conic Section
119844
The equation of the tangent to the circle \(x^2+y^2\) \(=25\) at \((4,3)\) is
1 \(4 x-3 y=25\)
2 \(4 x+3 y=25\)
3 \(4 x+3 y=126\)
4 \(4 x+3 y=9\)
Explanation:
B Given, \(x^2+y^2=25\) Point \((4,3)\) The slope of tangent is \(2 x+2 y \frac{d y}{d x}=0\) \(2 x=-2 y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{-x}{y}\) \(\left(\frac{d y}{d x}\right)_{(4,3)}=\frac{-4}{3}\) \(\therefore\) Equation of tangent is \(y-y_1=\left(\frac{d y}{d x}\right)_{(a, b)}\left(x-x_1\right)\) \((y-3)=\frac{-4}{3}(x-4)\) \(3 y-9=-4 x+16\) \(4 x+3 y=25\)
JEEE-2018
Conic Section
119845
A tangent is drawn to the circle \(2 x^2+2 y^2-3 x+4 y=0\) at the point ' \(A\) ' and it meets the line \(x+y=3\) at \(B(2,1)\), then \(A B=\)
1 \(\sqrt{10}\)
2 2
3 \(2 \sqrt{2}\)
4 0
Explanation:
B Equation of circle is, \(2 x^2+2 y^2-3 x+4 y=0\) \(x^2+y^2-\frac{3}{2} x+2 y=0\) Here, \(A B\) is the length of tangent to the circle from \(B\) \(\text { i.e. } A B=\sqrt{(2)^2+(1)^2-\frac{3}{2}(2)+2(1)}\) \(=\sqrt{4+1-3+2}=\sqrt{4}\) \(=2 \text {. }\)
Karnataka CET-2013
Conic Section
119852
The sum of the minimum distance and the maximum distance from the point \((4,-3)\) to the circle \(x^2+y^2+4 x-10 y-7=0\) is
1 20
2 12
3 10
4 16
Explanation:
A Centre of the given circle \(=\mathrm{C}(-2,5)\) Radius of the circle \(\mathrm{CN}=\mathrm{CT}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{2^2+5^2+7}=\sqrt{36}=6\) Distance between (4,-3) and ( \(-2,5\) ) \(\mathrm{PC}=\sqrt{6^2+8^2}=\sqrt{100}=10\) We join the external point, \((4,-3)\) to the centre of the circle \((-2,5)\). Then PT is the minimum distance, from external point \(\mathrm{P}\) to the circle and \(\mathrm{PN}\) is the maximum distance. Minimum distance \(=\mathrm{PT}=\mathrm{PC}-\mathrm{CT}=10-6=4\) Maximum distance \(=\mathrm{PN}=\mathrm{PC}+\mathrm{CN}=10+6=16\) So, sum of minimum and maximum distance \(=\) \(16+4=20\)
COMEDK-2018
Conic Section
119846
The equations of the two tangents from \((-5,-4)\) to the circle \(x^2+y^2+4 x+6 y+8=0\) are
