NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
120018
Let \(S_1 \Rightarrow x^2+y^2=9\) and \(S_2 \Rightarrow(x-2)^2+y^2=1\). Then the locus of center of a variable circle \(S_2\) which touches \(S_1\) internally and \(S_2\) externally always passes through the points
C Given, circle equation \(S_1=x^2+y^2=9\) \(c_1=(0,0), r_1=3\) \(S_2=(x-2)^2+y^2=1\) \(c_2=(2,0), r_2=1\) Let, centre of variable circle be \(c_3(h, k)\) and radius \(r\) \(c_3 c_1=3-\mathrm{r}\) \(\mathrm{c}_2 \mathrm{c}_3=1+\mathrm{r}\) \(\mathrm{c}_3 \mathrm{c}_1+\mathrm{c}_2 \mathrm{c}_3=4\) \(\sqrt{\mathrm{h}^2+\mathrm{k}^2}+\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4\) \(\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4-\sqrt{\mathrm{h}^2+\mathrm{k}^2} .\) \((\mathrm{h}-2)^2+\mathrm{k}^2=16+\mathrm{h}^2+\mathrm{k}^2-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(-4 \mathrm{~h}+4=16-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}+3=2 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}^2+6 \mathrm{~h}+9=4 \mathrm{~h}^2+4 \mathrm{k}^2\) \(3(\mathrm{~h}-1)^2+4 \mathrm{k}^2=12\) From the given option \(\left(2, \pm \frac{3}{2}\right) \text { satisfies it }\)
JEE Main 18.03.2021
Conic Section
120019
Choose the correct statement about two circles whose equations are given below. \(x^2+y^2-10 x-10 y+41=0\) \(x^2+y^2-22 x-10 y+137=0\)
1 circle have same centre
2 circles have no meeting point
3 circles have only one meeting point
4 circles have two meeting points
Explanation:
C Given, Two circle equation \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+41=0\) \((x-5)^2+(y-5)^2=3^2\) Centre, \(c_1=(5,5)\) and radius \(r_1=3\) And, \(S_2=x^2+y^2-22 x-10 y+137=0\) \((x-11)^2+(y-5)^2=3^2\) Centre, \(\mathrm{c}_2(11,5)\) and radius \(\mathrm{r}_2=3\) Now, \(c_1 c_2=\sqrt{(11-5)^2+(5-5)^2}=6\) And, \(r_1+r_2=3+3=6\) Since, \(\mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_1+\mathrm{r}_2=6\) So, Circle touch externally. Thus, the circle has only one meeting point.
JEE Main 18.03.2021
Conic Section
120020
If the circle (x+a)^2+(y+b)^2=a^2 \text { and }(x+c)^2+\) \((y+d)^2=d^2 \text { cuts orthogonally, then } b(b-2 d)=\)
1 \(c(c-2 a)\)
2 \(c(2 a-c)\)
3 \(\mathrm{d}(2 \mathrm{c}-\mathrm{a})\)
4 \(a(a-2 c)\)
Explanation:
B Given, two circle equation\(S_1=(x+a)^2+(y+b)^2=a^2\) \(\mathrm{c}_1=(-\mathrm{a},-\mathrm{b}) \text { and } \mathrm{r}_1=\mathrm{a}\) \(\mathrm{S}_2=(\mathrm{x}+\mathrm{c})^2+(\mathrm{y}+\mathrm{d})^2=\mathrm{d}^2\) \(\mathrm{c}_2=(-\mathrm{c},-\mathrm{d}) \text { and } \mathrm{r}_2=\mathrm{d}\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(a^2+d^2=a^2+c^2-2 a c+b^2+d^2-2 b d\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\) If two circle cut orthogonally then, \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}+\mathrm{b}^2+\mathrm{d}^2-2 \mathrm{bd}\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\)
AP EAMCET-23.04.2019
Conic Section
120021
Equation of circle passes through the points of \(\text { intersection of circles } x^2+y^2=6 \text { and } x^2+y^2-\) \(6 x+8=0 \text { and point }(1,1) \text { is }\)
1 \(x^2+y^2-6 x+4=0\)
2 \(x^2+y^2-3 x+1=0\)
3 \(x^2+y^2-4 y+2=0\)
4 \(x^2+y^2-6 x-6 y+10=0\)
Explanation:
B Given two circle equation, \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{S}_2=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8=0\) \(\mathrm{S}_1+\mathrm{k}\left(\mathrm{S}_2-\mathrm{S}_1\right)=0\) \(\mathrm{S}_1-\mathrm{S}_2=-6 \mathrm{x}+14\) \(\mathrm{x}^2+\mathrm{y}^2-6+\mathrm{k}(-6 \mathrm{x}+14)=0\) At point \((1,1)\) \(1^2+1^2-6+\mathrm{k}(-6(1)+14)=0\) \(-4+8 \mathrm{k}=0\) \(\mathrm{k}=\frac{1}{2}\) Now, \(2\left(x^2+y^2-6\right)+(1)(-6 x+14)=0\) \(2 x^2+2 y^2-12-6 x+14=0\) \(2 x^2+2 y^2-6 x+2=0\) \(x^2+y^2-3 x+1=0\)
120018
Let \(S_1 \Rightarrow x^2+y^2=9\) and \(S_2 \Rightarrow(x-2)^2+y^2=1\). Then the locus of center of a variable circle \(S_2\) which touches \(S_1\) internally and \(S_2\) externally always passes through the points
C Given, circle equation \(S_1=x^2+y^2=9\) \(c_1=(0,0), r_1=3\) \(S_2=(x-2)^2+y^2=1\) \(c_2=(2,0), r_2=1\) Let, centre of variable circle be \(c_3(h, k)\) and radius \(r\) \(c_3 c_1=3-\mathrm{r}\) \(\mathrm{c}_2 \mathrm{c}_3=1+\mathrm{r}\) \(\mathrm{c}_3 \mathrm{c}_1+\mathrm{c}_2 \mathrm{c}_3=4\) \(\sqrt{\mathrm{h}^2+\mathrm{k}^2}+\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4\) \(\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4-\sqrt{\mathrm{h}^2+\mathrm{k}^2} .\) \((\mathrm{h}-2)^2+\mathrm{k}^2=16+\mathrm{h}^2+\mathrm{k}^2-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(-4 \mathrm{~h}+4=16-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}+3=2 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}^2+6 \mathrm{~h}+9=4 \mathrm{~h}^2+4 \mathrm{k}^2\) \(3(\mathrm{~h}-1)^2+4 \mathrm{k}^2=12\) From the given option \(\left(2, \pm \frac{3}{2}\right) \text { satisfies it }\)
JEE Main 18.03.2021
Conic Section
120019
Choose the correct statement about two circles whose equations are given below. \(x^2+y^2-10 x-10 y+41=0\) \(x^2+y^2-22 x-10 y+137=0\)
1 circle have same centre
2 circles have no meeting point
3 circles have only one meeting point
4 circles have two meeting points
Explanation:
C Given, Two circle equation \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+41=0\) \((x-5)^2+(y-5)^2=3^2\) Centre, \(c_1=(5,5)\) and radius \(r_1=3\) And, \(S_2=x^2+y^2-22 x-10 y+137=0\) \((x-11)^2+(y-5)^2=3^2\) Centre, \(\mathrm{c}_2(11,5)\) and radius \(\mathrm{r}_2=3\) Now, \(c_1 c_2=\sqrt{(11-5)^2+(5-5)^2}=6\) And, \(r_1+r_2=3+3=6\) Since, \(\mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_1+\mathrm{r}_2=6\) So, Circle touch externally. Thus, the circle has only one meeting point.
JEE Main 18.03.2021
Conic Section
120020
If the circle (x+a)^2+(y+b)^2=a^2 \text { and }(x+c)^2+\) \((y+d)^2=d^2 \text { cuts orthogonally, then } b(b-2 d)=\)
1 \(c(c-2 a)\)
2 \(c(2 a-c)\)
3 \(\mathrm{d}(2 \mathrm{c}-\mathrm{a})\)
4 \(a(a-2 c)\)
Explanation:
B Given, two circle equation\(S_1=(x+a)^2+(y+b)^2=a^2\) \(\mathrm{c}_1=(-\mathrm{a},-\mathrm{b}) \text { and } \mathrm{r}_1=\mathrm{a}\) \(\mathrm{S}_2=(\mathrm{x}+\mathrm{c})^2+(\mathrm{y}+\mathrm{d})^2=\mathrm{d}^2\) \(\mathrm{c}_2=(-\mathrm{c},-\mathrm{d}) \text { and } \mathrm{r}_2=\mathrm{d}\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(a^2+d^2=a^2+c^2-2 a c+b^2+d^2-2 b d\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\) If two circle cut orthogonally then, \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}+\mathrm{b}^2+\mathrm{d}^2-2 \mathrm{bd}\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\)
AP EAMCET-23.04.2019
Conic Section
120021
Equation of circle passes through the points of \(\text { intersection of circles } x^2+y^2=6 \text { and } x^2+y^2-\) \(6 x+8=0 \text { and point }(1,1) \text { is }\)
1 \(x^2+y^2-6 x+4=0\)
2 \(x^2+y^2-3 x+1=0\)
3 \(x^2+y^2-4 y+2=0\)
4 \(x^2+y^2-6 x-6 y+10=0\)
Explanation:
B Given two circle equation, \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{S}_2=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8=0\) \(\mathrm{S}_1+\mathrm{k}\left(\mathrm{S}_2-\mathrm{S}_1\right)=0\) \(\mathrm{S}_1-\mathrm{S}_2=-6 \mathrm{x}+14\) \(\mathrm{x}^2+\mathrm{y}^2-6+\mathrm{k}(-6 \mathrm{x}+14)=0\) At point \((1,1)\) \(1^2+1^2-6+\mathrm{k}(-6(1)+14)=0\) \(-4+8 \mathrm{k}=0\) \(\mathrm{k}=\frac{1}{2}\) Now, \(2\left(x^2+y^2-6\right)+(1)(-6 x+14)=0\) \(2 x^2+2 y^2-12-6 x+14=0\) \(2 x^2+2 y^2-6 x+2=0\) \(x^2+y^2-3 x+1=0\)
120018
Let \(S_1 \Rightarrow x^2+y^2=9\) and \(S_2 \Rightarrow(x-2)^2+y^2=1\). Then the locus of center of a variable circle \(S_2\) which touches \(S_1\) internally and \(S_2\) externally always passes through the points
C Given, circle equation \(S_1=x^2+y^2=9\) \(c_1=(0,0), r_1=3\) \(S_2=(x-2)^2+y^2=1\) \(c_2=(2,0), r_2=1\) Let, centre of variable circle be \(c_3(h, k)\) and radius \(r\) \(c_3 c_1=3-\mathrm{r}\) \(\mathrm{c}_2 \mathrm{c}_3=1+\mathrm{r}\) \(\mathrm{c}_3 \mathrm{c}_1+\mathrm{c}_2 \mathrm{c}_3=4\) \(\sqrt{\mathrm{h}^2+\mathrm{k}^2}+\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4\) \(\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4-\sqrt{\mathrm{h}^2+\mathrm{k}^2} .\) \((\mathrm{h}-2)^2+\mathrm{k}^2=16+\mathrm{h}^2+\mathrm{k}^2-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(-4 \mathrm{~h}+4=16-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}+3=2 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}^2+6 \mathrm{~h}+9=4 \mathrm{~h}^2+4 \mathrm{k}^2\) \(3(\mathrm{~h}-1)^2+4 \mathrm{k}^2=12\) From the given option \(\left(2, \pm \frac{3}{2}\right) \text { satisfies it }\)
JEE Main 18.03.2021
Conic Section
120019
Choose the correct statement about two circles whose equations are given below. \(x^2+y^2-10 x-10 y+41=0\) \(x^2+y^2-22 x-10 y+137=0\)
1 circle have same centre
2 circles have no meeting point
3 circles have only one meeting point
4 circles have two meeting points
Explanation:
C Given, Two circle equation \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+41=0\) \((x-5)^2+(y-5)^2=3^2\) Centre, \(c_1=(5,5)\) and radius \(r_1=3\) And, \(S_2=x^2+y^2-22 x-10 y+137=0\) \((x-11)^2+(y-5)^2=3^2\) Centre, \(\mathrm{c}_2(11,5)\) and radius \(\mathrm{r}_2=3\) Now, \(c_1 c_2=\sqrt{(11-5)^2+(5-5)^2}=6\) And, \(r_1+r_2=3+3=6\) Since, \(\mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_1+\mathrm{r}_2=6\) So, Circle touch externally. Thus, the circle has only one meeting point.
JEE Main 18.03.2021
Conic Section
120020
If the circle (x+a)^2+(y+b)^2=a^2 \text { and }(x+c)^2+\) \((y+d)^2=d^2 \text { cuts orthogonally, then } b(b-2 d)=\)
1 \(c(c-2 a)\)
2 \(c(2 a-c)\)
3 \(\mathrm{d}(2 \mathrm{c}-\mathrm{a})\)
4 \(a(a-2 c)\)
Explanation:
B Given, two circle equation\(S_1=(x+a)^2+(y+b)^2=a^2\) \(\mathrm{c}_1=(-\mathrm{a},-\mathrm{b}) \text { and } \mathrm{r}_1=\mathrm{a}\) \(\mathrm{S}_2=(\mathrm{x}+\mathrm{c})^2+(\mathrm{y}+\mathrm{d})^2=\mathrm{d}^2\) \(\mathrm{c}_2=(-\mathrm{c},-\mathrm{d}) \text { and } \mathrm{r}_2=\mathrm{d}\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(a^2+d^2=a^2+c^2-2 a c+b^2+d^2-2 b d\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\) If two circle cut orthogonally then, \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}+\mathrm{b}^2+\mathrm{d}^2-2 \mathrm{bd}\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\)
AP EAMCET-23.04.2019
Conic Section
120021
Equation of circle passes through the points of \(\text { intersection of circles } x^2+y^2=6 \text { and } x^2+y^2-\) \(6 x+8=0 \text { and point }(1,1) \text { is }\)
1 \(x^2+y^2-6 x+4=0\)
2 \(x^2+y^2-3 x+1=0\)
3 \(x^2+y^2-4 y+2=0\)
4 \(x^2+y^2-6 x-6 y+10=0\)
Explanation:
B Given two circle equation, \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{S}_2=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8=0\) \(\mathrm{S}_1+\mathrm{k}\left(\mathrm{S}_2-\mathrm{S}_1\right)=0\) \(\mathrm{S}_1-\mathrm{S}_2=-6 \mathrm{x}+14\) \(\mathrm{x}^2+\mathrm{y}^2-6+\mathrm{k}(-6 \mathrm{x}+14)=0\) At point \((1,1)\) \(1^2+1^2-6+\mathrm{k}(-6(1)+14)=0\) \(-4+8 \mathrm{k}=0\) \(\mathrm{k}=\frac{1}{2}\) Now, \(2\left(x^2+y^2-6\right)+(1)(-6 x+14)=0\) \(2 x^2+2 y^2-12-6 x+14=0\) \(2 x^2+2 y^2-6 x+2=0\) \(x^2+y^2-3 x+1=0\)
120018
Let \(S_1 \Rightarrow x^2+y^2=9\) and \(S_2 \Rightarrow(x-2)^2+y^2=1\). Then the locus of center of a variable circle \(S_2\) which touches \(S_1\) internally and \(S_2\) externally always passes through the points
C Given, circle equation \(S_1=x^2+y^2=9\) \(c_1=(0,0), r_1=3\) \(S_2=(x-2)^2+y^2=1\) \(c_2=(2,0), r_2=1\) Let, centre of variable circle be \(c_3(h, k)\) and radius \(r\) \(c_3 c_1=3-\mathrm{r}\) \(\mathrm{c}_2 \mathrm{c}_3=1+\mathrm{r}\) \(\mathrm{c}_3 \mathrm{c}_1+\mathrm{c}_2 \mathrm{c}_3=4\) \(\sqrt{\mathrm{h}^2+\mathrm{k}^2}+\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4\) \(\sqrt{(\mathrm{h}-2)^2+\mathrm{k}^2}=4-\sqrt{\mathrm{h}^2+\mathrm{k}^2} .\) \((\mathrm{h}-2)^2+\mathrm{k}^2=16+\mathrm{h}^2+\mathrm{k}^2-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(-4 \mathrm{~h}+4=16-8 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}+3=2 \sqrt{\mathrm{h}^2+\mathrm{k}^2}\) \(\mathrm{~h}^2+6 \mathrm{~h}+9=4 \mathrm{~h}^2+4 \mathrm{k}^2\) \(3(\mathrm{~h}-1)^2+4 \mathrm{k}^2=12\) From the given option \(\left(2, \pm \frac{3}{2}\right) \text { satisfies it }\)
JEE Main 18.03.2021
Conic Section
120019
Choose the correct statement about two circles whose equations are given below. \(x^2+y^2-10 x-10 y+41=0\) \(x^2+y^2-22 x-10 y+137=0\)
1 circle have same centre
2 circles have no meeting point
3 circles have only one meeting point
4 circles have two meeting points
Explanation:
C Given, Two circle equation \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+41=0\) \((x-5)^2+(y-5)^2=3^2\) Centre, \(c_1=(5,5)\) and radius \(r_1=3\) And, \(S_2=x^2+y^2-22 x-10 y+137=0\) \((x-11)^2+(y-5)^2=3^2\) Centre, \(\mathrm{c}_2(11,5)\) and radius \(\mathrm{r}_2=3\) Now, \(c_1 c_2=\sqrt{(11-5)^2+(5-5)^2}=6\) And, \(r_1+r_2=3+3=6\) Since, \(\mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_1+\mathrm{r}_2=6\) So, Circle touch externally. Thus, the circle has only one meeting point.
JEE Main 18.03.2021
Conic Section
120020
If the circle (x+a)^2+(y+b)^2=a^2 \text { and }(x+c)^2+\) \((y+d)^2=d^2 \text { cuts orthogonally, then } b(b-2 d)=\)
1 \(c(c-2 a)\)
2 \(c(2 a-c)\)
3 \(\mathrm{d}(2 \mathrm{c}-\mathrm{a})\)
4 \(a(a-2 c)\)
Explanation:
B Given, two circle equation\(S_1=(x+a)^2+(y+b)^2=a^2\) \(\mathrm{c}_1=(-\mathrm{a},-\mathrm{b}) \text { and } \mathrm{r}_1=\mathrm{a}\) \(\mathrm{S}_2=(\mathrm{x}+\mathrm{c})^2+(\mathrm{y}+\mathrm{d})^2=\mathrm{d}^2\) \(\mathrm{c}_2=(-\mathrm{c},-\mathrm{d}) \text { and } \mathrm{r}_2=\mathrm{d}\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(a^2+d^2=a^2+c^2-2 a c+b^2+d^2-2 b d\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\) If two circle cut orthogonally then, \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{c}_1 \mathrm{c}_2\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\left(\sqrt{(-\mathrm{a}+\mathrm{c})^2+(-\mathrm{b}+\mathrm{d})^2}\right)^2\) \(\mathrm{a}^2+\mathrm{d}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}+\mathrm{b}^2+\mathrm{d}^2-2 \mathrm{bd}\) \(\mathrm{c}^2+\mathrm{b}^2-2 \mathrm{ac}-2 \mathrm{bd}=0\) \(\mathrm{~b}(\mathrm{~b}-2 \mathrm{~d})=-\mathrm{c}^2+2 \mathrm{ac}\) \(\mathrm{b}(\mathrm{b}-2 \mathrm{~d})=(2 \mathrm{a}-\mathrm{c}) \mathrm{c}\)
AP EAMCET-23.04.2019
Conic Section
120021
Equation of circle passes through the points of \(\text { intersection of circles } x^2+y^2=6 \text { and } x^2+y^2-\) \(6 x+8=0 \text { and point }(1,1) \text { is }\)
1 \(x^2+y^2-6 x+4=0\)
2 \(x^2+y^2-3 x+1=0\)
3 \(x^2+y^2-4 y+2=0\)
4 \(x^2+y^2-6 x-6 y+10=0\)
Explanation:
B Given two circle equation, \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{S}_2=\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+8=0\) \(\mathrm{S}_1+\mathrm{k}\left(\mathrm{S}_2-\mathrm{S}_1\right)=0\) \(\mathrm{S}_1-\mathrm{S}_2=-6 \mathrm{x}+14\) \(\mathrm{x}^2+\mathrm{y}^2-6+\mathrm{k}(-6 \mathrm{x}+14)=0\) At point \((1,1)\) \(1^2+1^2-6+\mathrm{k}(-6(1)+14)=0\) \(-4+8 \mathrm{k}=0\) \(\mathrm{k}=\frac{1}{2}\) Now, \(2\left(x^2+y^2-6\right)+(1)(-6 x+14)=0\) \(2 x^2+2 y^2-12-6 x+14=0\) \(2 x^2+2 y^2-6 x+2=0\) \(x^2+y^2-3 x+1=0\)
