119699
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
1 \(x^2+y^2-4 x-6 y+4=0\)
2 \(x^2+y^2-4 x-6 y-8=0\)
3 \(x^2+y^2-4 x+6 y+4=0\)
4 \(x^2+y^2+4 x-6 y+8=0\)
Explanation:
C Given centre (2,-3) Equation of circle, \(x^2+y^2+2 g x+2 f y+c=0\) and, centre \(\rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(x^2+y^2-4 x+6 y+c=0\) touching \(\mathrm{x}\) axis then \(\mathrm{y}=0\) \(4-8+c=0\) \(c=4\) Putting the value \(\mathrm{c}\) in equation \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}+4=0\)
AP EAMCET-07.07.2022
Conic Section
119701
For all real values of \(k\), the polar of the point \((2 k, k-4)\) with respect to \(x^2+y^2-4 x-6 y+1=\) 0 passes through the point
1 \((1,1)\)
2 \((1,-1)\)
3 \((-3,1)\)
4 \((3,1)\)
Explanation:
D Given equation is - \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}+1=0\) Equation of polar is \(\mathrm{T}=0\) \(\mathrm{xx}_1+\mathrm{yy}_1-4\left(\frac{\mathrm{x}+\mathrm{x}_1}{2}\right)-6\left(\frac{\mathrm{y}+\mathrm{y}_1}{2}\right)+1=0\) \(\mathrm{x}(2 \mathrm{k})+\mathrm{y}(\mathrm{k}-4)-2(\mathrm{x}+2 \mathrm{k})-3(\mathrm{y}+\mathrm{k}-4)+1=0\) \(2 \mathrm{kx}+\mathrm{yk}-4 \mathrm{y}-2 \mathrm{x}-4 \mathrm{k}-3 \mathrm{y}-3 \mathrm{k}+12+1=0\) \(\mathrm{k}(2 \mathrm{x}+\mathrm{y}-7)-(2 \mathrm{x}+7 \mathrm{y}-13)=0\) The equation of polar will be independent of \(k\) if its coefficient is zero. \(2 x+y-7=0\) \(2 x+7 y-13=0\) \(-\quad+\) \(-6 y+6=0\) \(y=1\) \(x=3\)So polar passes through the fix point \((3,1)\)
TS EAMCET-2016
Conic Section
119702
The value of a, such that the power of the point \((1,6)\) with respect to the circle \(x^2+y^2+4 x-6 y\) \(-\mathbf{a}=\mathbf{0}\), is \(-\mathbf{1 6}\) is,
1 7
2 11
3 13
4 21
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}-\mathrm{a}=0\) We know that power of point \(p(1,6)\) with respect the circle \(x^2+y^2+4 x-6 y-a=0\) is -16 Then, \((1)^2+(6)^2+4 \times 1-6 \times 6-a=-16\) \(1+36+4-36-a=-16\) \(\mathrm{a}=16+5\) \(a=21\)
TS EAMCET-2015
Conic Section
119703
The radius of the circle passing through the points \((-1,1),(2,-1)\) and \((1,0)\) is
1 5
2 \(\frac{\sqrt{130}}{2}\)
3 6
4 \(\frac{\sqrt{145}}{2}\)
Explanation:
B Given point \((-1,1)(2,-1),(1,0)\) General equation of circle is - \(x^2+y^2+2 g x+2 f y+c=0\) At point \((-1,1)\) \((1)^2+(1)^2+2 \mathrm{~g}(-1)+2 \mathrm{f}(1)+\mathrm{c}=0\) \(-2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0\) At point \((2,-1)\) \((2)^2+(-1)^2+2 \mathrm{~g}(2)+2 \mathrm{f}(-1)+\mathrm{c}=0\) \(4 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}+5=0\) By (i) and (ii), we get - \(2 \mathrm{~g}+2 \mathrm{c}+7=0\) At point \((1,0)\) \((1)^2+0+2 \mathrm{~g}+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) By solving equation (iii) and (iv) we get - \(c=-6, \quad g=5 / 2\) By putting the value of \(\mathrm{c} \& \mathrm{~g}\) in equation (i) we get - \(\mathrm{f}=9 / 2\) Now, Radius of the circle is - \(r=\sqrt{g^2+f^2-c}\) \(=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{9}{2}\right)^2+6}=\frac{\sqrt{130}}{2}\)
TS EAMCET-19.07.2022
Conic Section
119704
If the circle \(x^2+y^2-6 x+2 y=28\) cuts off a chord of length \(\lambda\) units on the line \(2 x-5 y+18\) \(=0\), then the value of \(\lambda\) is
1 3
2 6
3 12
4 9
Explanation:
B Given circle equation \(x^2+y^2-6 x+2 y=28\) by comparing general equation of circle we get \(\mathrm{g}=-3 \quad \mathrm{f}=1, \quad \mathrm{c}=-28\) \(\text { Radius }(\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(1)^2-(-28)}=\sqrt{38}\) \(\mathrm{CP}=\frac{2 \times 3-5 \times(-1)+18}{\sqrt{(2)^2+(-5)^2}}\) \(\mathrm{CP}=\frac{29}{\sqrt{29}}=\sqrt{29}\) \(\text { From } \Delta \mathrm{CPA}\) \(\mathrm{AC}^2=\mathrm{CP}^2+(\mathrm{AP})^2\) \((\sqrt{38})^2=(\sqrt{29})^2+(\mathrm{AP})^2\) \(38=29+(\mathrm{AP})^2\) \((\mathrm{AP})^2=38-29=9\) \(\mathrm{AP}=3\)Lenth of chord of the circle \(=2 \times \mathrm{AP}=2 \times 3=6\)
119699
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
1 \(x^2+y^2-4 x-6 y+4=0\)
2 \(x^2+y^2-4 x-6 y-8=0\)
3 \(x^2+y^2-4 x+6 y+4=0\)
4 \(x^2+y^2+4 x-6 y+8=0\)
Explanation:
C Given centre (2,-3) Equation of circle, \(x^2+y^2+2 g x+2 f y+c=0\) and, centre \(\rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(x^2+y^2-4 x+6 y+c=0\) touching \(\mathrm{x}\) axis then \(\mathrm{y}=0\) \(4-8+c=0\) \(c=4\) Putting the value \(\mathrm{c}\) in equation \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}+4=0\)
AP EAMCET-07.07.2022
Conic Section
119701
For all real values of \(k\), the polar of the point \((2 k, k-4)\) with respect to \(x^2+y^2-4 x-6 y+1=\) 0 passes through the point
1 \((1,1)\)
2 \((1,-1)\)
3 \((-3,1)\)
4 \((3,1)\)
Explanation:
D Given equation is - \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}+1=0\) Equation of polar is \(\mathrm{T}=0\) \(\mathrm{xx}_1+\mathrm{yy}_1-4\left(\frac{\mathrm{x}+\mathrm{x}_1}{2}\right)-6\left(\frac{\mathrm{y}+\mathrm{y}_1}{2}\right)+1=0\) \(\mathrm{x}(2 \mathrm{k})+\mathrm{y}(\mathrm{k}-4)-2(\mathrm{x}+2 \mathrm{k})-3(\mathrm{y}+\mathrm{k}-4)+1=0\) \(2 \mathrm{kx}+\mathrm{yk}-4 \mathrm{y}-2 \mathrm{x}-4 \mathrm{k}-3 \mathrm{y}-3 \mathrm{k}+12+1=0\) \(\mathrm{k}(2 \mathrm{x}+\mathrm{y}-7)-(2 \mathrm{x}+7 \mathrm{y}-13)=0\) The equation of polar will be independent of \(k\) if its coefficient is zero. \(2 x+y-7=0\) \(2 x+7 y-13=0\) \(-\quad+\) \(-6 y+6=0\) \(y=1\) \(x=3\)So polar passes through the fix point \((3,1)\)
TS EAMCET-2016
Conic Section
119702
The value of a, such that the power of the point \((1,6)\) with respect to the circle \(x^2+y^2+4 x-6 y\) \(-\mathbf{a}=\mathbf{0}\), is \(-\mathbf{1 6}\) is,
1 7
2 11
3 13
4 21
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}-\mathrm{a}=0\) We know that power of point \(p(1,6)\) with respect the circle \(x^2+y^2+4 x-6 y-a=0\) is -16 Then, \((1)^2+(6)^2+4 \times 1-6 \times 6-a=-16\) \(1+36+4-36-a=-16\) \(\mathrm{a}=16+5\) \(a=21\)
TS EAMCET-2015
Conic Section
119703
The radius of the circle passing through the points \((-1,1),(2,-1)\) and \((1,0)\) is
1 5
2 \(\frac{\sqrt{130}}{2}\)
3 6
4 \(\frac{\sqrt{145}}{2}\)
Explanation:
B Given point \((-1,1)(2,-1),(1,0)\) General equation of circle is - \(x^2+y^2+2 g x+2 f y+c=0\) At point \((-1,1)\) \((1)^2+(1)^2+2 \mathrm{~g}(-1)+2 \mathrm{f}(1)+\mathrm{c}=0\) \(-2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0\) At point \((2,-1)\) \((2)^2+(-1)^2+2 \mathrm{~g}(2)+2 \mathrm{f}(-1)+\mathrm{c}=0\) \(4 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}+5=0\) By (i) and (ii), we get - \(2 \mathrm{~g}+2 \mathrm{c}+7=0\) At point \((1,0)\) \((1)^2+0+2 \mathrm{~g}+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) By solving equation (iii) and (iv) we get - \(c=-6, \quad g=5 / 2\) By putting the value of \(\mathrm{c} \& \mathrm{~g}\) in equation (i) we get - \(\mathrm{f}=9 / 2\) Now, Radius of the circle is - \(r=\sqrt{g^2+f^2-c}\) \(=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{9}{2}\right)^2+6}=\frac{\sqrt{130}}{2}\)
TS EAMCET-19.07.2022
Conic Section
119704
If the circle \(x^2+y^2-6 x+2 y=28\) cuts off a chord of length \(\lambda\) units on the line \(2 x-5 y+18\) \(=0\), then the value of \(\lambda\) is
1 3
2 6
3 12
4 9
Explanation:
B Given circle equation \(x^2+y^2-6 x+2 y=28\) by comparing general equation of circle we get \(\mathrm{g}=-3 \quad \mathrm{f}=1, \quad \mathrm{c}=-28\) \(\text { Radius }(\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(1)^2-(-28)}=\sqrt{38}\) \(\mathrm{CP}=\frac{2 \times 3-5 \times(-1)+18}{\sqrt{(2)^2+(-5)^2}}\) \(\mathrm{CP}=\frac{29}{\sqrt{29}}=\sqrt{29}\) \(\text { From } \Delta \mathrm{CPA}\) \(\mathrm{AC}^2=\mathrm{CP}^2+(\mathrm{AP})^2\) \((\sqrt{38})^2=(\sqrt{29})^2+(\mathrm{AP})^2\) \(38=29+(\mathrm{AP})^2\) \((\mathrm{AP})^2=38-29=9\) \(\mathrm{AP}=3\)Lenth of chord of the circle \(=2 \times \mathrm{AP}=2 \times 3=6\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119699
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
1 \(x^2+y^2-4 x-6 y+4=0\)
2 \(x^2+y^2-4 x-6 y-8=0\)
3 \(x^2+y^2-4 x+6 y+4=0\)
4 \(x^2+y^2+4 x-6 y+8=0\)
Explanation:
C Given centre (2,-3) Equation of circle, \(x^2+y^2+2 g x+2 f y+c=0\) and, centre \(\rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(x^2+y^2-4 x+6 y+c=0\) touching \(\mathrm{x}\) axis then \(\mathrm{y}=0\) \(4-8+c=0\) \(c=4\) Putting the value \(\mathrm{c}\) in equation \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}+4=0\)
AP EAMCET-07.07.2022
Conic Section
119701
For all real values of \(k\), the polar of the point \((2 k, k-4)\) with respect to \(x^2+y^2-4 x-6 y+1=\) 0 passes through the point
1 \((1,1)\)
2 \((1,-1)\)
3 \((-3,1)\)
4 \((3,1)\)
Explanation:
D Given equation is - \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}+1=0\) Equation of polar is \(\mathrm{T}=0\) \(\mathrm{xx}_1+\mathrm{yy}_1-4\left(\frac{\mathrm{x}+\mathrm{x}_1}{2}\right)-6\left(\frac{\mathrm{y}+\mathrm{y}_1}{2}\right)+1=0\) \(\mathrm{x}(2 \mathrm{k})+\mathrm{y}(\mathrm{k}-4)-2(\mathrm{x}+2 \mathrm{k})-3(\mathrm{y}+\mathrm{k}-4)+1=0\) \(2 \mathrm{kx}+\mathrm{yk}-4 \mathrm{y}-2 \mathrm{x}-4 \mathrm{k}-3 \mathrm{y}-3 \mathrm{k}+12+1=0\) \(\mathrm{k}(2 \mathrm{x}+\mathrm{y}-7)-(2 \mathrm{x}+7 \mathrm{y}-13)=0\) The equation of polar will be independent of \(k\) if its coefficient is zero. \(2 x+y-7=0\) \(2 x+7 y-13=0\) \(-\quad+\) \(-6 y+6=0\) \(y=1\) \(x=3\)So polar passes through the fix point \((3,1)\)
TS EAMCET-2016
Conic Section
119702
The value of a, such that the power of the point \((1,6)\) with respect to the circle \(x^2+y^2+4 x-6 y\) \(-\mathbf{a}=\mathbf{0}\), is \(-\mathbf{1 6}\) is,
1 7
2 11
3 13
4 21
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}-\mathrm{a}=0\) We know that power of point \(p(1,6)\) with respect the circle \(x^2+y^2+4 x-6 y-a=0\) is -16 Then, \((1)^2+(6)^2+4 \times 1-6 \times 6-a=-16\) \(1+36+4-36-a=-16\) \(\mathrm{a}=16+5\) \(a=21\)
TS EAMCET-2015
Conic Section
119703
The radius of the circle passing through the points \((-1,1),(2,-1)\) and \((1,0)\) is
1 5
2 \(\frac{\sqrt{130}}{2}\)
3 6
4 \(\frac{\sqrt{145}}{2}\)
Explanation:
B Given point \((-1,1)(2,-1),(1,0)\) General equation of circle is - \(x^2+y^2+2 g x+2 f y+c=0\) At point \((-1,1)\) \((1)^2+(1)^2+2 \mathrm{~g}(-1)+2 \mathrm{f}(1)+\mathrm{c}=0\) \(-2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0\) At point \((2,-1)\) \((2)^2+(-1)^2+2 \mathrm{~g}(2)+2 \mathrm{f}(-1)+\mathrm{c}=0\) \(4 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}+5=0\) By (i) and (ii), we get - \(2 \mathrm{~g}+2 \mathrm{c}+7=0\) At point \((1,0)\) \((1)^2+0+2 \mathrm{~g}+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) By solving equation (iii) and (iv) we get - \(c=-6, \quad g=5 / 2\) By putting the value of \(\mathrm{c} \& \mathrm{~g}\) in equation (i) we get - \(\mathrm{f}=9 / 2\) Now, Radius of the circle is - \(r=\sqrt{g^2+f^2-c}\) \(=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{9}{2}\right)^2+6}=\frac{\sqrt{130}}{2}\)
TS EAMCET-19.07.2022
Conic Section
119704
If the circle \(x^2+y^2-6 x+2 y=28\) cuts off a chord of length \(\lambda\) units on the line \(2 x-5 y+18\) \(=0\), then the value of \(\lambda\) is
1 3
2 6
3 12
4 9
Explanation:
B Given circle equation \(x^2+y^2-6 x+2 y=28\) by comparing general equation of circle we get \(\mathrm{g}=-3 \quad \mathrm{f}=1, \quad \mathrm{c}=-28\) \(\text { Radius }(\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(1)^2-(-28)}=\sqrt{38}\) \(\mathrm{CP}=\frac{2 \times 3-5 \times(-1)+18}{\sqrt{(2)^2+(-5)^2}}\) \(\mathrm{CP}=\frac{29}{\sqrt{29}}=\sqrt{29}\) \(\text { From } \Delta \mathrm{CPA}\) \(\mathrm{AC}^2=\mathrm{CP}^2+(\mathrm{AP})^2\) \((\sqrt{38})^2=(\sqrt{29})^2+(\mathrm{AP})^2\) \(38=29+(\mathrm{AP})^2\) \((\mathrm{AP})^2=38-29=9\) \(\mathrm{AP}=3\)Lenth of chord of the circle \(=2 \times \mathrm{AP}=2 \times 3=6\)
119699
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
1 \(x^2+y^2-4 x-6 y+4=0\)
2 \(x^2+y^2-4 x-6 y-8=0\)
3 \(x^2+y^2-4 x+6 y+4=0\)
4 \(x^2+y^2+4 x-6 y+8=0\)
Explanation:
C Given centre (2,-3) Equation of circle, \(x^2+y^2+2 g x+2 f y+c=0\) and, centre \(\rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(x^2+y^2-4 x+6 y+c=0\) touching \(\mathrm{x}\) axis then \(\mathrm{y}=0\) \(4-8+c=0\) \(c=4\) Putting the value \(\mathrm{c}\) in equation \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}+4=0\)
AP EAMCET-07.07.2022
Conic Section
119701
For all real values of \(k\), the polar of the point \((2 k, k-4)\) with respect to \(x^2+y^2-4 x-6 y+1=\) 0 passes through the point
1 \((1,1)\)
2 \((1,-1)\)
3 \((-3,1)\)
4 \((3,1)\)
Explanation:
D Given equation is - \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}+1=0\) Equation of polar is \(\mathrm{T}=0\) \(\mathrm{xx}_1+\mathrm{yy}_1-4\left(\frac{\mathrm{x}+\mathrm{x}_1}{2}\right)-6\left(\frac{\mathrm{y}+\mathrm{y}_1}{2}\right)+1=0\) \(\mathrm{x}(2 \mathrm{k})+\mathrm{y}(\mathrm{k}-4)-2(\mathrm{x}+2 \mathrm{k})-3(\mathrm{y}+\mathrm{k}-4)+1=0\) \(2 \mathrm{kx}+\mathrm{yk}-4 \mathrm{y}-2 \mathrm{x}-4 \mathrm{k}-3 \mathrm{y}-3 \mathrm{k}+12+1=0\) \(\mathrm{k}(2 \mathrm{x}+\mathrm{y}-7)-(2 \mathrm{x}+7 \mathrm{y}-13)=0\) The equation of polar will be independent of \(k\) if its coefficient is zero. \(2 x+y-7=0\) \(2 x+7 y-13=0\) \(-\quad+\) \(-6 y+6=0\) \(y=1\) \(x=3\)So polar passes through the fix point \((3,1)\)
TS EAMCET-2016
Conic Section
119702
The value of a, such that the power of the point \((1,6)\) with respect to the circle \(x^2+y^2+4 x-6 y\) \(-\mathbf{a}=\mathbf{0}\), is \(-\mathbf{1 6}\) is,
1 7
2 11
3 13
4 21
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}-\mathrm{a}=0\) We know that power of point \(p(1,6)\) with respect the circle \(x^2+y^2+4 x-6 y-a=0\) is -16 Then, \((1)^2+(6)^2+4 \times 1-6 \times 6-a=-16\) \(1+36+4-36-a=-16\) \(\mathrm{a}=16+5\) \(a=21\)
TS EAMCET-2015
Conic Section
119703
The radius of the circle passing through the points \((-1,1),(2,-1)\) and \((1,0)\) is
1 5
2 \(\frac{\sqrt{130}}{2}\)
3 6
4 \(\frac{\sqrt{145}}{2}\)
Explanation:
B Given point \((-1,1)(2,-1),(1,0)\) General equation of circle is - \(x^2+y^2+2 g x+2 f y+c=0\) At point \((-1,1)\) \((1)^2+(1)^2+2 \mathrm{~g}(-1)+2 \mathrm{f}(1)+\mathrm{c}=0\) \(-2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0\) At point \((2,-1)\) \((2)^2+(-1)^2+2 \mathrm{~g}(2)+2 \mathrm{f}(-1)+\mathrm{c}=0\) \(4 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}+5=0\) By (i) and (ii), we get - \(2 \mathrm{~g}+2 \mathrm{c}+7=0\) At point \((1,0)\) \((1)^2+0+2 \mathrm{~g}+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) By solving equation (iii) and (iv) we get - \(c=-6, \quad g=5 / 2\) By putting the value of \(\mathrm{c} \& \mathrm{~g}\) in equation (i) we get - \(\mathrm{f}=9 / 2\) Now, Radius of the circle is - \(r=\sqrt{g^2+f^2-c}\) \(=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{9}{2}\right)^2+6}=\frac{\sqrt{130}}{2}\)
TS EAMCET-19.07.2022
Conic Section
119704
If the circle \(x^2+y^2-6 x+2 y=28\) cuts off a chord of length \(\lambda\) units on the line \(2 x-5 y+18\) \(=0\), then the value of \(\lambda\) is
1 3
2 6
3 12
4 9
Explanation:
B Given circle equation \(x^2+y^2-6 x+2 y=28\) by comparing general equation of circle we get \(\mathrm{g}=-3 \quad \mathrm{f}=1, \quad \mathrm{c}=-28\) \(\text { Radius }(\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(1)^2-(-28)}=\sqrt{38}\) \(\mathrm{CP}=\frac{2 \times 3-5 \times(-1)+18}{\sqrt{(2)^2+(-5)^2}}\) \(\mathrm{CP}=\frac{29}{\sqrt{29}}=\sqrt{29}\) \(\text { From } \Delta \mathrm{CPA}\) \(\mathrm{AC}^2=\mathrm{CP}^2+(\mathrm{AP})^2\) \((\sqrt{38})^2=(\sqrt{29})^2+(\mathrm{AP})^2\) \(38=29+(\mathrm{AP})^2\) \((\mathrm{AP})^2=38-29=9\) \(\mathrm{AP}=3\)Lenth of chord of the circle \(=2 \times \mathrm{AP}=2 \times 3=6\)
119699
The equation of circle with centre \((2,-3)\) and touching \(x\)-axis is
1 \(x^2+y^2-4 x-6 y+4=0\)
2 \(x^2+y^2-4 x-6 y-8=0\)
3 \(x^2+y^2-4 x+6 y+4=0\)
4 \(x^2+y^2+4 x-6 y+8=0\)
Explanation:
C Given centre (2,-3) Equation of circle, \(x^2+y^2+2 g x+2 f y+c=0\) and, centre \(\rightarrow(-\mathrm{g},-\mathrm{f})\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(x^2+y^2-4 x+6 y+c=0\) touching \(\mathrm{x}\) axis then \(\mathrm{y}=0\) \(4-8+c=0\) \(c=4\) Putting the value \(\mathrm{c}\) in equation \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+6 \mathrm{y}+4=0\)
AP EAMCET-07.07.2022
Conic Section
119701
For all real values of \(k\), the polar of the point \((2 k, k-4)\) with respect to \(x^2+y^2-4 x-6 y+1=\) 0 passes through the point
1 \((1,1)\)
2 \((1,-1)\)
3 \((-3,1)\)
4 \((3,1)\)
Explanation:
D Given equation is - \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}+1=0\) Equation of polar is \(\mathrm{T}=0\) \(\mathrm{xx}_1+\mathrm{yy}_1-4\left(\frac{\mathrm{x}+\mathrm{x}_1}{2}\right)-6\left(\frac{\mathrm{y}+\mathrm{y}_1}{2}\right)+1=0\) \(\mathrm{x}(2 \mathrm{k})+\mathrm{y}(\mathrm{k}-4)-2(\mathrm{x}+2 \mathrm{k})-3(\mathrm{y}+\mathrm{k}-4)+1=0\) \(2 \mathrm{kx}+\mathrm{yk}-4 \mathrm{y}-2 \mathrm{x}-4 \mathrm{k}-3 \mathrm{y}-3 \mathrm{k}+12+1=0\) \(\mathrm{k}(2 \mathrm{x}+\mathrm{y}-7)-(2 \mathrm{x}+7 \mathrm{y}-13)=0\) The equation of polar will be independent of \(k\) if its coefficient is zero. \(2 x+y-7=0\) \(2 x+7 y-13=0\) \(-\quad+\) \(-6 y+6=0\) \(y=1\) \(x=3\)So polar passes through the fix point \((3,1)\)
TS EAMCET-2016
Conic Section
119702
The value of a, such that the power of the point \((1,6)\) with respect to the circle \(x^2+y^2+4 x-6 y\) \(-\mathbf{a}=\mathbf{0}\), is \(-\mathbf{1 6}\) is,
1 7
2 11
3 13
4 21
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}-6 \mathrm{y}-\mathrm{a}=0\) We know that power of point \(p(1,6)\) with respect the circle \(x^2+y^2+4 x-6 y-a=0\) is -16 Then, \((1)^2+(6)^2+4 \times 1-6 \times 6-a=-16\) \(1+36+4-36-a=-16\) \(\mathrm{a}=16+5\) \(a=21\)
TS EAMCET-2015
Conic Section
119703
The radius of the circle passing through the points \((-1,1),(2,-1)\) and \((1,0)\) is
1 5
2 \(\frac{\sqrt{130}}{2}\)
3 6
4 \(\frac{\sqrt{145}}{2}\)
Explanation:
B Given point \((-1,1)(2,-1),(1,0)\) General equation of circle is - \(x^2+y^2+2 g x+2 f y+c=0\) At point \((-1,1)\) \((1)^2+(1)^2+2 \mathrm{~g}(-1)+2 \mathrm{f}(1)+\mathrm{c}=0\) \(-2 \mathrm{~g}+2 \mathrm{f}+\mathrm{c}+2=0\) At point \((2,-1)\) \((2)^2+(-1)^2+2 \mathrm{~g}(2)+2 \mathrm{f}(-1)+\mathrm{c}=0\) \(4 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}+5=0\) By (i) and (ii), we get - \(2 \mathrm{~g}+2 \mathrm{c}+7=0\) At point \((1,0)\) \((1)^2+0+2 \mathrm{~g}+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) By solving equation (iii) and (iv) we get - \(c=-6, \quad g=5 / 2\) By putting the value of \(\mathrm{c} \& \mathrm{~g}\) in equation (i) we get - \(\mathrm{f}=9 / 2\) Now, Radius of the circle is - \(r=\sqrt{g^2+f^2-c}\) \(=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{9}{2}\right)^2+6}=\frac{\sqrt{130}}{2}\)
TS EAMCET-19.07.2022
Conic Section
119704
If the circle \(x^2+y^2-6 x+2 y=28\) cuts off a chord of length \(\lambda\) units on the line \(2 x-5 y+18\) \(=0\), then the value of \(\lambda\) is
1 3
2 6
3 12
4 9
Explanation:
B Given circle equation \(x^2+y^2-6 x+2 y=28\) by comparing general equation of circle we get \(\mathrm{g}=-3 \quad \mathrm{f}=1, \quad \mathrm{c}=-28\) \(\text { Radius }(\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(1)^2-(-28)}=\sqrt{38}\) \(\mathrm{CP}=\frac{2 \times 3-5 \times(-1)+18}{\sqrt{(2)^2+(-5)^2}}\) \(\mathrm{CP}=\frac{29}{\sqrt{29}}=\sqrt{29}\) \(\text { From } \Delta \mathrm{CPA}\) \(\mathrm{AC}^2=\mathrm{CP}^2+(\mathrm{AP})^2\) \((\sqrt{38})^2=(\sqrt{29})^2+(\mathrm{AP})^2\) \(38=29+(\mathrm{AP})^2\) \((\mathrm{AP})^2=38-29=9\) \(\mathrm{AP}=3\)Lenth of chord of the circle \(=2 \times \mathrm{AP}=2 \times 3=6\)
